tsc2001 yeah I used this and now I shoot 94 % from the free throw line which if you do not play basketball is a 15 foot shot that an average pro player hits at an efficiency of 75 %
It's not sad at all, that's how humans have been able to develop such incredible things. It's because of the relationship between mathematics and nature, otherwise we wouldn't be able to discover so many things, there is no other way, we wouldn't know anything without mathematics! :D@@yusufchaudhary6354
@@ayshaghanawi9941 I'm actually doing pretty well. Graduated high school and university with a chemistry degree, took couple physics and math classes and loved them all. I have been working as a chemist for 2 years now.
@@tunaovoxo Congrats! That sounds great! God bless you. Hope you continue with a successful future. This just comes to show us how fast time passes and how fast things can change. Glad you're doing well. It made me feel hopeful :)
let me summarize this. (khan does not use the most direct way of solving this, but its kind of cool nonetheless). We are given that the initial velocity of a particle has a magnitude of 10 meters/second, and points at an angle of 30 degrees with the horizontal to the right (so this is a vector in 2 dimensions, and we don't need a third dimension). We want to find the total distance this particle covers under only the influence of gravity (no air resistance involved). First thing we want to do is break down the velocity vector *v* into horizontal and vertical components. That way we can recast a 2-d problem into two 1-d problems, since we know our motion equations for 1-d motion. So again, we are given that || *v* ||= 10, and theta = 30 degrees. So we can use our knowledge of trigonometry. v_x = || *v* || cos theta v_y = || *v* || sin theta v_x = 10 cos 30 = 8.66 m/s v_y = 10 sin 30 = 5 m/s Now lets look at the vertical and horizontal motion separately. We know that delta v_y = delta t, where means average acceleration in the vertical direction (not instantaneous acceleration). But we know that for vertical motion in free fall, the acceleration due to gravity is a constant -9.8 m/s^2. Therefore we can substitute the constant 'a' of -9.8 for . And we also know that v_yi = 5 m/s and v_yf = -5 m/s by symmetry (sal explains it in the video). v_yf - v_yi = a* delta t (-5) - 5 = -9.8* delta t So delta t = 10/9.8 = 1.02 seconds, which is how long the particle is in the air. Now we also know that delta s_x = *delta t. But here v_x is constant so = v_x , since gravity has no effect on an objects horizontal motion, only vertical. So plugging in we have delta s_x = 8.66 * delta t delta s_x = 8.66 * 1.02 = 8.8 meters.
Physics is interesting, but it can be a difficult thing to understand and rationalize. A lot of physics is hard to grasp because its so different from what you perceived things to be. The math also, can be a little tough.
Oh my god I struggled so much in physics doing calculations for stuff I had no idea why I was doing that and in the first minute you already explained 1 million things I was never told! Why we separate the vertical and hor. velocity and how we use them to calculate what. You can only do physics if you understand what you are doing and I can imagine how many people you guys have saved with your videos. thank you
Sal, you are so amazing! Its appreciable how one person can explain so many subjects in detail!! Videos by Khan Academy have helped me in understanding a lot of stuff and will always help me in the future! Thank You!!
I used equations of uniformly accelerated motion in order to solve it. The one I used was hight = 1/2 g t^2 where t is the time from the velocity equal to zero to the final velocity before it smacks the ground. The total time period was 1.02 sec so the half will be 0.51 sec which is what I plugged in the equation which is two times less than your answer why is that the case? When I plugged the whole time period then it's in line with yours. Shouldn't the half of the time used in the equation?
At 6:25 when he said that -1*(initial velocity)= final velocity I actually got a more mathematical proof of that statement through contradiction and differential calculus. It's too long but if you want it reply to this comment
What I did was section the object at different places, that is: Pt. A = starting scenario of the projectile Pt. B = where the projectile is at its peak height i.e. (zero vertical velocity) Pt. C = ending point of the projectile Similarly, I broke the vector down into its x & y components (Note: the x and the y motion of the object is independent and are examined separately USING 1D KINEMATICS EQNS BUT they do share a similar the same parameter which is TIME) THE SOLUTION: 1. The motion of the object from A to B is SIMILAR to its motion from B to C. So I only examined the motion from A to B. 2. I listed the parameters available in both horizontal and vertical motion (from Pt. A to B) of the projectile. Horizontal Parameters available u = 10Cos30 = 5√3 a = 0 (since horizontal velocity is constant for projectile motion) Vertical Parameters available u = 10Sin30 = 5 a = -9.8 (negative since the vertical velocity from A to B is decreasing...at Pt. B vertical velocity is zero as learned in 1D kinematics taught by Sir Mahesh Shenoy) v = 0 (as said above, the final velocity we're considering here is at Pt. B i.e. at max height of projectile where vertical velocity is zero) 3. Pick the appropriate 1D kinematics eqn for each horizontal and vertical motion to solve for the time, t, since it is the parameter common to both horizontal and vertical motion. (time is usually solved in vertical motion since vertical motion offers a lot more parameters to work with) Vertical motion from Pt. A to Pt. B: v = u + at 0 = 5 + (-9.8)t t = 0.51 Horizontal motion from Pt. A to Pt. B: s = ut + (1/2)*at^2 s = (5√3)*t + 0 s = (5√3)*0.51 ≈ 4.42 (BUT this is incomplete, since this is only the horizontal distance from A to B. we need to multiply the time by 2, therefore doubling the s as expected) Hence Final Answer is s ≈ 8.83 (Note: the time it took for the projectile to travel from A to B is the same as from B to C, AND since the horizontal distance we need is from A to C, it is in our intuition to double the initial computed s from A to B as a result of doubling the computed time in the vertical motion i.e. from A to B)
OR you can use the Range Equation. its a whole lot faster and its applicable due to y's displacement being 0. R= (Vo^2*sin2theta)/g R=((10^2)(sin(60)))/9.8 = 8.83m
if you want the range just use this formula R=vo^2(sin2@)\g . @= means the theta angle or whatever angle name it it the fastest way to solve the questing just use the formula
@khanacademy hi sal , so i have a question about the vertical velocity , correct me if i was wrong so to get the normal average velocity we do it like this ''delta''V =(Vi+Vf)/2 , but here when you calculated the vertical velocity which is also ''delta''V you calculated it like this ''delta''V = (Vf-Vi) and you used the same formula for the average velocity which is: a=''delta''V/''delta''t so whtat's the difference between those two velocities ?
I assume that you have got the calculation part and are not able to understand logically. Well, here the initial velocity is the velocity just after the moment the rocket was fired. Like how the initial velocity of a bullet will be the velocity of the bullet just after it was fired. Similarly, the final velocity will be the velocity of the object just before it hits the ground, not the velocity after it hits the ground. The magnitude will be same as initial velocity as the elevation of the ground is not changing and since we are taking up direction as +ve, the final velocity will be -ve as the direction of the velocity is downwards (as the object is falling).
As it goes upwards the object slows down (acceleration is negative) As it reaches its peak point its velocity is 0. As it accelerates downwards, it gains velocity but in the opposite direction to where it first started (opposite of upward is downward) So it gains negative velocity.
Let u be the initial velocity and v the final. Assume -1*u≠v Take derivative on both sides to get acceleration. So, -1*d/dx(u)≠d/dx(v) This is not true! We are working under a constant acceleration which is 9.8 m/s² hence our hypothesis is wrong. Hence -1*(initial velocity)= final velocity. Multiplication by -1 due to the convention of up being positive and down being negative.
This is only true for the vertical component of velocity i.e. v_y = -v0_y . The horizontal component v_x = v0_x as there is no air drag. The magnitude of the initial and final velocity is thereby the same sqrt(v_x^2 + v_y^2) = sqrt(v_0x^2 + v_0y^2)
@@78anurag If you multiply the initial velocity vector by (-1), you are effectively reflecting it about both the horizontal and vertical axes. While the magnitude will still be correct, the horizontal direction is not correct, as there is no air drag.
@@drofeng This proof does not apply to horizontal velocity yes, because in horizontal velocity -1*(initial velocity)≠final velocity. But it does apply to the projectile velocity and vertical component of velocity
Somebody please tell me why the final velocity of vertical component has the same magnitude as the initial velocity of the vertical component? I thought it should be different
Because the deceleration (first half of the motion) and acceleration (second half of the motion) due to gravity is the same but opposite. Hence, v_y = -v_0y and v_x = v_0x
So how would I calculate the angle of initial projectile and angle of impact on a parabolic trajectory? If I were given a ballistics chart based on known drag coefficient specs? For example trajectory leaves at 290 feet traveling a distance of 2000 feet hitting an object at 30 feet from the ground...
when you start calculating your change in vertical velocity why doesn't your equation have the initial velocity added in on the right side. I am also confused how the initial velocity is 10m/s when in reality isn't it always zero?
Why can we use 10m/s as a length for the hypotenuse when its just speed in a given direction? I know its a vector quantity but it doesnt quite make sense to me..
Shaon Kabir because that's the resultant vector (so to speak). So the resultant vector velocity is 10ms^-1 , therefore we use that as the hypotenuse (we need the Vx and Vy components)
I feel like if physicists can calculate gravity, they should be able to settle on some sort of constant for air resistance. Most of these problems are not valid in the real world.
So given the known distance of the target we need to work backward on the angle with the fixed launch velocity of the projectile. How did the pre-satellite artillery men determine the distance between them and the target before they fired?
sal calculated the time it takes to go up to max height amounting to about 1 sec. however, the time taken to reach the ground is twice the time taken to reach max height this means time of flight is 2 seconds so the distance should be twice the value calculated i.e 17.6 m please explain if I'm right or where i've gone wrong
It seemed a very counterintuitive answer that it should only go 8 meters when 10m/s launch speed is like 36kmph and I was thinking if one threw a ball at that speed surely as we all know it would go way further . But I was thinking that analogy isn’t the same because even if you threw the ball at the same 30 degree angle you’d be throwing it from a height well above the ground so that added height you’ve given it initially clearly increases it’s air time and total distance . This example would have the Launcher literally sitting on the ground . An average throw speed for a ball would be much faster anyway I’d say .
+Mark Henry, the initial velocity here is when any sort of short-acting acceleration finishes launching the projectile. So if you were throwing a rock, it would be right when you let go of the rock -- thereafter it is a problem with a known velocity and direction and a known acceleration downward.
I a bit confused about the triangle in the video. If the hypotenuse it 10 and it is a right triangle then shouldn't the sides fulfill the 3:4:5 ratio? If the angle is 30deg and it is a right triangle then shouldn't it fulfill the 30,60,90 right triangle pattern? It seems like the two patterns are mutually exclusive...I get a different (incorrect?) answer if I solve the problem using the 3:4:5 ratio. can anyone help me reconcile this?
I don´t understand how he is able to know for certain the final vertical velocity of -5 m/s. As I understand, at that point he has too many unknown variables in the acceleration equations to solve for final vertical velocity.
The thing is air resistance depend of many things like the shape of the objet for example, so it can't be a constant. Plus when you fall you gain speed and the faster you are the more air resistance you get, so you'll have to use a differential equation. They are ways to calculate it but it's kind of complex and take a lot of paradigms so physic problems in school usually don't bother with it.
why not to know the hang time by that rule y=(viy)(t)-1/2 (g)(t)power 2 like for the first example we have 10 and 30○ :- 0=10sin(30)(x)-1/2 (9.8)(x)power 2 and shift solve on the calculator =1.0204 as we know everything expect the t that we already trying to get g=is always 9.8
Who else is watching this in 6th grade? Just me? So far I understand nothing. Ooh they are explaining it as if they knew someone like me would watch. Never-mind I regret trying to calculate projectile motion in 6th grade.
my 4 year old bought me here. He's dead set about trajectories... when racing hot wheel in midair... I think i need to get him a physics & math tutor. What 4 year old does that?
I'm so impressed that Sal can teach so many subjects so well
This really makes me understand how science and maths suddenly become relevant to each other.
Cool! Wait till you get to *Relativity* ;)
Wait, sir when will we ever use trig?
When using 2d projectile motion at an angle of course!
tsc2001 yeah I used this and now I shoot 94 % from the free throw line which if you do not play basketball is a 15 foot shot that an average pro player hits at an efficiency of 75 %
Sadly
It's not sad at all, that's how humans have been able to develop such incredible things. It's because of the relationship between mathematics and nature, otherwise we wouldn't be able to discover so many things, there is no other way, we wouldn't know anything without mathematics! :D@@yusufchaudhary6354
Sal Khan is the greatest gift to us mortals. We are unworthy of his teaching effectiveness.
bro calm down he's not a god.
man you're the reason why i have a 90 in physics. thank you
How are you doing now? It's been 7 years
@@ayshaghanawi9941 I'm actually doing pretty well. Graduated high school and university with a chemistry degree, took couple physics and math classes and loved them all. I have been working as a chemist for 2 years now.
@@tunaovoxo Woah congrats
@@annika7594 thanks!!
@@tunaovoxo Congrats! That sounds great! God bless you. Hope you continue with a successful future. This just comes to show us how fast time passes and how fast things can change. Glad you're doing well. It made me feel hopeful :)
let me summarize this. (khan does not use the most direct way of solving this, but its kind of cool nonetheless).
We are given that the initial velocity of a particle has a magnitude of 10 meters/second, and points at an angle of 30 degrees with the horizontal to the right (so this is a vector in 2 dimensions, and we don't need a third dimension). We want to find the total distance this particle covers under only the influence of gravity (no air resistance involved).
First thing we want to do is break down the velocity vector *v* into horizontal and vertical components. That way we can recast a 2-d problem into two 1-d problems, since we know our motion equations for 1-d motion.
So again, we are given that || *v* ||= 10, and theta = 30 degrees. So we can use our knowledge of trigonometry.
v_x = || *v* || cos theta
v_y = || *v* || sin theta
v_x = 10 cos 30 = 8.66 m/s
v_y = 10 sin 30 = 5 m/s
Now lets look at the vertical and horizontal motion separately.
We know that
delta v_y = delta t, where means average acceleration in the vertical direction (not instantaneous acceleration). But we know that for vertical motion in free fall, the acceleration due to gravity is a constant -9.8 m/s^2.
Therefore we can substitute the constant 'a' of -9.8 for .
And we also know that v_yi = 5 m/s and v_yf = -5 m/s by symmetry (sal explains it in the video).
v_yf - v_yi = a* delta t
(-5) - 5 = -9.8* delta t
So delta t = 10/9.8 = 1.02 seconds, which is how long the particle is in the air.
Now we also know that
delta s_x = *delta t.
But here v_x is constant so = v_x , since gravity has no effect on an objects horizontal motion, only vertical. So plugging in we have
delta s_x = 8.66 * delta t
delta s_x = 8.66 * 1.02 = 8.8 meters.
thanks alot! it was a blast learning this even though it's far beyond my school level
physics is hard
it is
is it hard if you understand it?
No one said it wasn't interesting.
Physics is interesting, but it can be a difficult thing to understand and rationalize. A lot of physics is hard to grasp because its so different from what you perceived things to be. The math also, can be a little tough.
Nothing good in life is easy
Oh my god I struggled so much in physics doing calculations for stuff I had no idea why I was doing that and in the first minute you already explained 1 million things I was never told! Why we separate the vertical and hor. velocity and how we use them to calculate what. You can only do physics if you understand what you are doing and I can imagine how many people you guys have saved with your videos. thank you
I did not expect the giant calculator lol
"we can assume that we are doing this experiment on the moon" , Still uses the Earth's gravitational force :)
Exactly my thought! Ugh, my physics teacher would be so proud.
There is a gravitational force on Moon that's less than 9.8 m/s^2.
@@jenniferratto9232 i thought it’d 9.81
PEDANTRYYYY!!!!! GRRR
he was just assuming.... not thinking about gravity but rather about the atmosphere and air
i could learn more and faster if i just followed my curriculum with his videos...id actually learn something
Sal, you are so amazing! Its appreciable how one person can explain so many subjects in detail!! Videos by Khan Academy have helped me in understanding a lot of stuff and will always help me in the future! Thank You!!
Thank god for this video; everything makes so much more sense now :o
All 35 dislikes are by people who make similar videos but charge for it 😂
If anyone touches his like button you will be cursed for making it 70
@@user-vv5ju3fj1p i am scared
I make similar videos and don't charge for it, and no ads :) Might help ^^
WOW! THAT WAS AMAZING!!
Man you are the reason why i came to know that physics is this hard
I used equations of uniformly accelerated motion in order to solve it. The one I used was hight = 1/2 g t^2 where t is the time from the velocity equal to zero to the final velocity before it smacks the ground. The total time period was 1.02 sec so the half will be 0.51 sec which is what I plugged in the equation which is two times less than your answer why is that the case? When I plugged the whole time period then it's in line with yours. Shouldn't the half of the time used in the equation?
im gonna end up working at mcdonalds
Haha! Ah....... :,(
At 6:25 when he said that -1*(initial velocity)= final velocity I actually got a more mathematical proof of that statement through contradiction and differential calculus. It's too long but if you want it reply to this comment
Go ahead! I can't make it make sense.
I wish I had found it before taking my exams, educational nonetheless! :) Great video. Looking forward to learning a lot!
What I did was section the object at different places, that is:
Pt. A = starting scenario of the projectile
Pt. B = where the projectile is at its peak height i.e. (zero vertical velocity)
Pt. C = ending point of the projectile
Similarly, I broke the vector down into its x & y components (Note: the x and the y motion of the object is independent and are examined separately USING 1D KINEMATICS EQNS BUT they do share a similar the same parameter which is TIME)
THE SOLUTION:
1. The motion of the object from A to B is SIMILAR to its motion from B to C. So I only examined the motion from A to B.
2. I listed the parameters available in both horizontal and vertical motion (from Pt. A to B) of the projectile.
Horizontal Parameters available
u = 10Cos30 = 5√3
a = 0 (since horizontal velocity is constant for projectile motion)
Vertical Parameters available
u = 10Sin30 = 5
a = -9.8 (negative since the vertical velocity from A to B is decreasing...at Pt. B vertical velocity is zero as learned in 1D kinematics taught by Sir Mahesh Shenoy)
v = 0 (as said above, the final velocity we're considering here is at Pt. B i.e. at max height of projectile where vertical velocity is zero)
3. Pick the appropriate 1D kinematics eqn for each horizontal and vertical motion to solve for the time, t, since it is the parameter common to both horizontal and vertical motion. (time is usually solved in vertical motion since vertical motion offers a lot more parameters to work with)
Vertical motion from Pt. A to Pt. B:
v = u + at
0 = 5 + (-9.8)t
t = 0.51
Horizontal motion from Pt. A to Pt. B:
s = ut + (1/2)*at^2
s = (5√3)*t + 0
s = (5√3)*0.51 ≈ 4.42 (BUT this is incomplete, since this is only the horizontal distance from A to B. we need to multiply the time by 2, therefore doubling the s as expected)
Hence Final Answer is s ≈ 8.83
(Note: the time it took for the projectile to travel from A to B is the same as from B to C, AND since the horizontal distance we need is from A to C, it is in our intuition to double the initial computed s from A to B as a result of doubling the computed time in the vertical motion i.e. from A to B)
OR you can use the Range Equation. its a whole lot faster and its applicable due to y's displacement being 0.
R= (Vo^2*sin2theta)/g R=((10^2)(sin(60)))/9.8 = 8.83m
Da faq
Incredible video. So simply explained in a manner that is short and to the point.
I have an exam in 15 minutes, you’re a lifesaver 😨😨
I have a question about gravity in 7:35, why did you put negative gravity (-9,8) and no positive gravity (+9,8) ? Thank you in advance!!
Gravity is going down so it’s negative
if you want the range just use this formula R=vo^2(sin2@)\g .
@= means the theta angle or whatever angle name
it it the fastest way to solve the questing
just use the formula
By the way, an easy way to fin Range is:
Range=sin(2(theta))*(velocity)^2/(gravitational acceleration)
How old are you now man?
Range=V^2.sin(2Q)/g = use this equation to find horizontal displacement
you guys are my hero
really ur helping me a lot.....right now..esp.. wen i dont get concepts in scool...
@khanacademy
hi sal ,
so i have a question about the vertical velocity , correct me if i was wrong so to get the normal average velocity we do it like this ''delta''V =(Vi+Vf)/2 , but here when you calculated the vertical velocity which is also ''delta''V you calculated it like this ''delta''V = (Vf-Vi) and you used the same formula for the average velocity which is: a=''delta''V/''delta''t so whtat's the difference between those two velocities ?
very good tutorial...you are doing a very good work!!
how do you write so good with your mouse?
when i try, i write HRLO instead of HELLO
He is not using a mouse to write. He uses a drawing tablet like from Wacom.
How did you get initial velocity to = 5 m/s and final velocity to equal -5 m/s? That is where I'm lost. In my head, initial is 0 and so is final.
I assume that you have got the calculation part and are not able to understand logically.
Well, here the initial velocity is the velocity just after the moment the rocket was fired. Like how the initial velocity of a bullet will be the velocity of the bullet just after it was fired.
Similarly, the final velocity will be the velocity of the object just before it hits the ground, not the velocity after it hits the ground. The magnitude will be same as initial velocity as the elevation of the ground is not changing and since we are taking up direction as +ve, the final velocity will be -ve as the direction of the velocity is downwards (as the object is falling).
As it goes upwards the object slows down (acceleration is negative)
As it reaches its peak point its velocity is 0.
As it accelerates downwards, it gains velocity but in the opposite direction to where it first started (opposite of upward is downward)
So it gains negative velocity.
Let u be the initial velocity and v the final. Assume -1*u≠v
Take derivative on both sides to get acceleration.
So,
-1*d/dx(u)≠d/dx(v)
This is not true! We are working under a constant acceleration which is 9.8 m/s² hence our hypothesis is wrong.
Hence -1*(initial velocity)= final velocity.
Multiplication by -1 due to the convention of up being positive and down being negative.
This is only true for the vertical component of velocity i.e. v_y = -v0_y . The horizontal component v_x = v0_x as there is no air drag. The magnitude of the initial and final velocity is thereby the same sqrt(v_x^2 + v_y^2) = sqrt(v_0x^2 + v_0y^2)
@@drofeng But I'm still correct since I'm talking about the projectile velocity and not certain components of the velocity right?
@@78anurag If you multiply the initial velocity vector by (-1), you are effectively reflecting it about both the horizontal and vertical axes. While the magnitude will still be correct, the horizontal direction is not correct, as there is no air drag.
@@drofeng This proof does not apply to horizontal velocity yes, because in horizontal velocity -1*(initial velocity)≠final velocity.
But it does apply to the projectile velocity and vertical component of velocity
Air resistance is dependent on many factors including the temperature, speed and the object's drag coefficient.
wait if there is acceleration due to gravity in the "y" then how is the initial and final velocity the same magnitude?
Veronica Marie if the elevation stays the same, the object will fall down just as fast as it was sent up.
you saved my life thanks
How do u know which to use when solvingT~T (soh-cah-toa)
thanks a lot...i just got the idea
How is the initial velocity and final velocity in vertical direction same in magnitude?
Is it because the initial and final position is the same?
Somebody please tell me why the final velocity of vertical component has the same magnitude as the initial velocity of the vertical component? I thought it should be different
Because the deceleration (first half of the motion) and acceleration (second half of the motion) due to gravity is the same but opposite. Hence, v_y = -v_0y and v_x = v_0x
I dont understand the way you did the change in x= 10cos30 = 5Sqroot of 3, when I do that on the calculator I get 8.66 not 10. I that right or wrong?
feeling so crazy that i used the same method u used to remember the trigonometry equations
So how would I calculate the angle of initial projectile and angle of impact on a parabolic trajectory? If I were given a ballistics chart based on known drag coefficient specs? For example trajectory leaves at 290 feet traveling a distance of 2000 feet hitting an object at 30 feet from the ground...
It's so different to what I'm trying to figure out.
This video is very helpful❤️
one question, wont g be negative when y direction moves up and then positive when it moves down? you cant take it to be -ve for both the sides!!
Misbah Mashkoor -5m/s (final verlocity) - (-5m/s) (initial velocity
=-10m/s
when you start calculating your change in vertical velocity why doesn't your equation have the initial velocity added in on the right side. I am also confused how the initial velocity is 10m/s when in reality isn't it always zero?
can someone explain to me why he chose to use -9.81 (at 7:53) and not positive 9.81? :(
The negative just indicates direction! Since the launched rocket is falling downwards you would make it -9.81
@@kevinzheng5091 correct whether the object goes up or down gravitational acceleration is always taken to be negative
How in the world does he get those little formulas? I know Vf=vi+at but where is he getting those delta from?
+Phoenix Floyd Delta just means change, so change in time for example.
The equations used are the equations of uniform motion.
Why can we use 10m/s as a length for the hypotenuse when its just speed in a given direction? I know its a vector quantity but it doesnt quite make sense to me..
Shaon Kabir because that's the resultant vector (so to speak). So the resultant vector velocity is 10ms^-1 , therefore we use that as the hypotenuse (we need the Vx and Vy components)
how did you get the initial velocity of 5m/s?
If it is projected at a velocity of 10m/s and it is in the air for 1.02 seconds, then shouldn't the displacement be at least 10 meters?
I feel like if physicists can calculate gravity, they should be able to settle on some sort of constant for air resistance. Most of these problems are not valid in the real world.
how would you calculate initial velocity if you know the launch angle and the time that its in the air?
This is very useful
Why is the final velocity -5m/s? Wouldn't it be 0m/s at it's final velocity once it hit's the ground?
So given the known distance of the target we need to work backward on the angle with the fixed launch velocity of the projectile. How did the pre-satellite artillery men determine the distance between them and the target before they fired?
Was this supposed to help me or question everything I have ever thought
Epic video!
The formula: dx = vi^2 sin(2angle) /g can be also used right? instead of dx = vix t?
How do i find the distance and initial velocity if only the angle and height is given?
thanks khan.
can you divide meters per second by meters per second squared?
sal calculated the time it takes to go up to max height amounting to about 1 sec.
however, the time taken to reach the ground is twice the time taken to reach max height
this means time of flight is 2 seconds
so the distance should be twice the value calculated i.e 17.6 m
please explain if I'm right or where i've gone wrong
NICE VID SAL
color transitioning is difficult. hahahaahah
It seemed a very counterintuitive answer that it should only go 8 meters when 10m/s launch speed is like 36kmph and I was thinking if one threw a ball at that speed surely as we all know it would go way further . But I was thinking that analogy isn’t the same because even if you threw the ball at the same 30 degree angle you’d be throwing it from a height well above the ground so that added height you’ve given it initially clearly increases it’s air time and total distance . This example would have the Launcher literally sitting on the ground . An average throw speed for a ball would be much faster anyway I’d say .
Is there a video somewhere on describing the flight of a rocket launched at an angle with constant thrust and linearly decreasing mass?
thank you so much for making these videos
hi can someone explain why the magnitude of the final velocity is equal to the initial velocity?
how did you find the time(delta t)
wouldnt the intial velocity be 0 m/s???
+Mark Henry, the initial velocity here is when any sort of short-acting acceleration finishes launching the projectile. So if you were throwing a rock, it would be right when you let go of the rock -- thereafter it is a problem with a known velocity and direction and a known acceleration downward.
can anyone tell me why sin30 = 1/2 in this problem?
Joel Mais that's the value of sin(30º) or sin(pi/6)
Joel Mais basic trig like he said
u got a long way in physics bud
the way he wrote meters a second made me think he was writing m/5.
I didn't understood how initial and final velocity are same, if some one knows please explain
Why would you even use sin, when we know, that in a right triangle if side is oposite to the 30 degree angle, it's half hypotenuse in size
What if you are only given the total displacement and the angle?
I a bit confused about the triangle in the video. If the hypotenuse it 10 and it is a right triangle then shouldn't the sides fulfill the 3:4:5 ratio? If the angle is 30deg and it is a right triangle then shouldn't it fulfill the 30,60,90 right triangle pattern? It seems like the two patterns are mutually exclusive...I get a different (incorrect?) answer if I solve the problem using the 3:4:5 ratio. can anyone help me reconcile this?
If you are still looking for the answer, I doubt you are, I used both methods and got the right answer, you may have just made a mistake somewhere
You could use Pythagoras theorem to get the Vx.
Your a lifesaver.
I don´t understand how he is able to know for certain the final vertical velocity of -5 m/s. As I understand, at that point he has too many unknown variables in the acceleration equations to solve for final vertical velocity.
How do we get-9.8 m/sec?
Saved my life
you're my king
Thank you so much.
I’d just aim shoot and hope for the best...
Same like if did the calculation and shot it it would end up at a worse aim then if I just aimed my self 😭💀
may GOd bless you
The thing is air resistance depend of many things like the shape of the objet for example, so it can't be a constant. Plus when you fall you gain speed and the faster you are the more air resistance you get, so you'll have to use a differential equation. They are ways to calculate it but it's kind of complex and take a lot of paradigms so physic problems in school usually don't bother with it.
My new man crush! thank u
@rdcruick watch the previous videos of projectile motion added by sal sir in the playlist physics it willl explain well
So good.
why not to know the hang time by that rule y=(viy)(t)-1/2 (g)(t)power 2 like for the first example we have 10 and 30○ :- 0=10sin(30)(x)-1/2 (9.8)(x)power 2 and shift solve on the calculator =1.0204 as we know everything expect the t that we already trying to get g=is always 9.8
nice
Who else is watching this in 6th grade?
Just me?
So far I understand nothing.
Ooh they are explaining it as if they knew someone like me would watch.
Never-mind I regret trying to calculate projectile motion in 6th grade.
I'm in 11th grade watching this so why are you?
time in the air is time up + time down i think you are missing something
what is a magniuted
What program do you use to make these videos? i've been really curious...
U going to copy him?
@@m.arhumirfan4570 huh
i still wonder how he films these videos
probably a tablet, stylus, and whatever program he uses (with a screen recorder) , along with a microphone to record. that's my guess tbh
my 4 year old bought me here. He's dead set about trajectories... when racing hot wheel in midair... I think i need to get him a physics & math tutor. What 4 year old does that?