I wish the questions I'm being asked were as straight forward as these are. They're complicated enough to make you think, but not easy enough to figure out without understanding the concepts. I'm so screwed in physics. The online homework is like harvard level questions, versus what the textbook teaches, and the professor explains. I wish I would have been set on the right path from the start. I'm at least 80 hours into trying to figure out all of this, and am still totally unsure of how to approach almost every problem I come across. Every single question has a curveball. Your videos are a light guiding me in the right direction tho, it's been dark af for a long time. Thank you for creating so many practice problems, with great explanations.
In this instance, if you find peak height you get something around 12m which isn't high enough to go over the wall. And that is the peak height meaning it never goes over. From that, you can infer the same that you got, no it does not clear the wall.
@Denseworldproduction Yes, that would essentially be the same problem. Do the horizontal part first, and find the time, then find the height *above* the starting point at that time and see if it will clear the wall.
Yes, your point is certainly valid. It would depend on the particular problem. There are situations where a 0.01 difference won't matter at all, and situations where it could meant the difference in success and total failure.
@WinsonLaw93 Remember that you have to treat the horizontal motion independently from the vertical motion. So we first find the time it takes to go *horizontally* to the wall. Then we find whether or not it is, at that moment in time, above the wall, vertically. Hope that helps, D.O.
It can be either way, depending on how you set up the problem. The key is to choose one particular direction, either up or down, to be positive, and then stick with that choice through the entire problem.
@WinsonLaw93 Good question. We can't use the peak height to determine if it will clear the wall because it is not at its peak height vertically at the same time it is horizontally at the wall. What he have to find is whether it is above the height of the wall at the moment that it is at the horizontal position of the wall. And that is a separate question from What is the peak height.
@Gsmstfreshie Not necessarily. The total range of the projectile, providing that there would be no wall in between, could be longer than 45m, which would also increase the total flight time. Now if this would be the total flight time, it would mean that the projectile hit the ground at the 45m. That is not the case, because we just calculated the time it takes the projectile to reach the wall.
You can calculate this the opposite way using the vertical component for a reference to the horizontal. To do this you would calculate the time that it takes for the rock to reach a height of 20 metres (on it's decent) and substitute that time into the horizontal component to find that the rock has only reached a horizontal distance of X (my estimate would be between 35-40 metres without actually calculating).
@albasser89 I think your formula will give you the total time, and if you use this time to find the horizontal distance, then you find that it does reach the wall. The question, though, is whether or not it is high enough to clear the top of the wall at the time that it reaches the wall, so you need to approach it differently.
great job! this helps me a lot, i used to be totally lost when doing cos, sin, and tan while finding ViY for trig, but this helped, THANKS! great drawings by the way.
Why can't we use the peak height to determine if the object will vault the castle? Why do we have to use the total time based off of the horizontal distance? By the way, I have a quantitative physics final exam this week, and this is really helping me! Even though my textbook (Giancolli 6th ed.) is a bit tougher, I am really getting the foundations I need to excel. Thank you so much! I appreciate your helping of students around the world!
My Mid year exams for physics are next week (Australian yr 12), and these help sooo much! i usually try to do them myself and then see if i get it right, but you have really helped me grasp the concepts, THANKS!
@kklola1 Yes, exactly! The physics in Angry Birds is pretty good. I think they calculate friction and momentum, too - just guessing based on how it all looks.
So I used [[[[[[[ v_f = v_0 + at ]]]]]] and I got t = 1.586 s, then I plugged it into [[[[[[ y_f = y_0 + v_0*t + ((1/2)at^2) ]]]]] and I got 12.34 m, which means the rock still did not go over the wall, but how to know which equation to use. x_f=vt or v_f = v_0 +at
Please can someone do this question as an example...(a car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal.The negligent driver leaves the car in neutral,and the emergency brakes are defective.The car rolls from rest down the incline with a constant acceleration of 4 meter per squared seconds for a distance of 50m to the edge of the cliff,which is 30m above the ocean. )......(a) find the length of time the car is in air.(b) find the car's position relative to the base of the cliff when the car lands in the ocean..
There are different ways to set up these problems. Your teacher may be doing it differently, but is certainly using a method that works out to be mathematically equivalent. The way I set these up is this: The x component of the velocity is found by multiplying v0 by the cosine of the angle; the y component of the velocity is found by multiplying v0 by the sine of the angle.
for anyone who still has that question.. you don't use the peak height because the peak height doesn't tell you the height at the time you're to the castle wall, which is all that matters.
@derekowens This is a really stupid question, but what do you mean by the moment at the horizontal position of the wall? Do you mean when the trajectory ends?
I did what you did initially, that is solve for t=x/v, but then i realise that the title says:" does it go over the wall?" so i used the v(peak)=v(sub o at the y direction) + at to find out the time of flight first. But then it turns out to be wrong
I have a question. Why did you use 2.89 s directly to solve for the height of the rock at that time? Wouldn't that be not correct? The rock will first go upward before going downward, so it will use different signs of gravitational acceleration. I'm confused.
i was thinking the same. He would have had to use the difference between the total time and the time it took to ascend. A projectile could also cross a certain point on its way down
can you show me how to do this problem if im given a height above ground as my start point, and angle but its asking the same question as above, how would i solve this? no velocity given
I think the answer is yes not negative. Because, The rock would go maximum 49 meters horizontally. It would also go 41 meters. So, it is actually that the rock would cross the wall.
The question, though, isn't simply about the maximum distance and the maximum height. The maximum distance may be more than 45, and the max height may be more than 20. The question though, is how high is it after it has traveled 45 meters horizontally. That is the question we need to answer to see if it will clear the wall.
For anyone who has seen the video recently.. For the distance, while calculating the time, he used 45 meters. That's the distance from the catapult to the wall, is it right just assume that's also the distance the rock travels (what he did in the video) although we haven't been given that? I started with the vertical component and used the Vf=Vi+at and got the time=1.6 seconds instead of what he got (2.89). Am I making a mistake here ?
Ahmad, 1.6 seconds is the correct value for the time to the peak, the highest point. That doesn't quite help us solve the problem, though. Using a time of 1.6 seconds we can find the height at the peak, and it turns out to be over 20 meters, but after the peak it starts coming back down, and we still don't know if it is above 20m at the time it reaches the wall. So there are two other ways we can solve it. We can find the time it takes to go horizontally 45 meters, and then see if the height is above 20m at that time. Or we can find the time at which it is 20m high, and then see if it is horizontally beyond 45 meters at that time. Either way will work, but starting by finding the time to go horizontally is a bit easier in this particular problem. Hope that helps!
not to be a super geek but if you keep the exact values through out the problem, you will get a little less than 4m like around 3.99m. I know it's irrelevant here but I think it can make a huge difference in real world no? like .2m can let the rock pass over the wall if wall was hmm lets say 3.9m tall.... However, great videossssss....i learned a lot from them as you can see :P
is my method correct? since the wall is 20 meters high, i just solve for the maximum height of the projectile which is 12.33 meters and just by seeing the maximum height I can assume that it will not pass through the wall.
Yes, that is valid, at least in *this* case. If the maximum height were higher than 20 meters, you would have to do a little more work. Specifically, you would need to determine if it is above 20m at the same moment it is at the horizontal position of the wall. Your approach shows a good intuition for solving this quickly. The more general solution is also valuable, though.
The acceleration is *down* simply because gravity pulls down. For any problem, you can decide if you want up to be the positive direction or down to be the positive direction. Either way is okay, just decide and stick to it through the end of the problem. If we make up the positive direction, then down is the negative direction, and since gravity pulls down the acceleration is therefore negative. Hope that helps!
In that case, you should be given some other information that will either allow you to find it, or will allow you to find something that will allow you to find it. In other words, from what you are given, you can typically find *something*, then something else, then something else, until you find the particular value you are aiming to find.
I could not get it. Why the vertical time equals to the horizontal time check it out at 5:40 ???? I think we have to divide by 2 to find the time vertical. Please answer my question ,,,
I'll try to help. From one point along to the trajectory to a second point along the trajectory will take a certain time. We can consider the horizontal motion during this time or the vertical motion during this time, but the time should be the same in each case. There are problems where we need to consider half of the flight versus the whole flight, and in those problems we often end up multiplying or dividing by 2, but this problem is different. Here, we are finding the time for the rock to reach the wall, which is generally not going to be at the midpoint of the flight.
R should be 45m. or higher R = Vo²sin2(¢)/g = (22)²sin(2×45)/(9.81) = 49.34 m. H should be 20m. or higher H = Vo²sin²(¢)/2g = (22)²sin²(45)/2(9.81) = 12.33 m. yes, it will reach??? 😂
I wish the questions I'm being asked were as straight forward as these are. They're complicated enough to make you think, but not easy enough to figure out without understanding the concepts. I'm so screwed in physics. The online homework is like harvard level questions, versus what the textbook teaches, and the professor explains. I wish I would have been set on the right path from the start. I'm at least 80 hours into trying to figure out all of this, and am still totally unsure of how to approach almost every problem I come across. Every single question has a curveball. Your videos are a light guiding me in the right direction tho, it's been dark af for a long time. Thank you for creating so many practice problems, with great explanations.
In this instance, if you find peak height you get something around 12m which isn't high enough to go over the wall. And that is the peak height meaning it never goes over. From that, you can infer the same that you got, no it does not clear the wall.
well drawn, well explained, clean, concise, no time wasted or distractions in the explanation. Good example. Thanks.
@Denseworldproduction Yes, that would essentially be the same problem. Do the horizontal part first, and find the time, then find the height *above* the starting point at that time and see if it will clear the wall.
Thanks for these videos. They're helping me a lot in drilling for an upcoming quiz.
Yes, your point is certainly valid. It would depend on the particular problem. There are situations where a 0.01 difference won't matter at all, and situations where it could meant the difference in success and total failure.
@WinsonLaw93 Remember that you have to treat the horizontal motion independently from the vertical motion. So we first find the time it takes to go *horizontally* to the wall. Then we find whether or not it is, at that moment in time, above the wall, vertically. Hope that helps, D.O.
It can be either way, depending on how you set up the problem. The key is to choose one particular direction, either up or down, to be positive, and then stick with that choice through the entire problem.
@WinsonLaw93 Good question. We can't use the peak height to determine if it will clear the wall because it is not at its peak height vertically at the same time it is horizontally at the wall. What he have to find is whether it is above the height of the wall at the moment that it is at the horizontal position of the wall. And that is a separate question from What is the peak height.
this has helped me for my project, many thanks
@Gsmstfreshie
Not necessarily. The total range of the projectile, providing that there would be no wall in between, could be longer than 45m, which would also increase the total flight time.
Now if this would be the total flight time, it would mean that the projectile hit the ground at the 45m.
That is not the case, because we just calculated the time it takes the projectile to reach the wall.
You can calculate this the opposite way using the vertical component for a reference to the horizontal.
To do this you would calculate the time that it takes for the rock to reach a height of 20 metres (on it's decent) and substitute that time into the horizontal component to find that the rock has only reached a horizontal distance of X (my estimate would be between 35-40 metres without actually calculating).
@albasser89 I think your formula will give you the total time, and if you use this time to find the horizontal distance, then you find that it does reach the wall. The question, though, is whether or not it is high enough to clear the top of the wall at the time that it reaches the wall, so you need to approach it differently.
Thank you very much! really helped me out
You saved me in calculus based physics
you are amazing!!!! Please make videos on the chapter uncertainties if possible
great job! this helps me a lot, i used to be totally lost when doing cos, sin, and tan while finding ViY for trig, but this helped, THANKS! great drawings by the way.
A 10 year old comment!
Why can't we use the peak height to determine if the object will vault the castle? Why do we have to use the total time based off of the horizontal distance?
By the way, I have a quantitative physics final exam this week, and this is really helping me! Even though my textbook (Giancolli 6th ed.) is a bit tougher, I am really getting the foundations I need to excel. Thank you so much! I appreciate your helping of students around the world!
Great question!!
My Mid year exams for physics are next week (Australian yr 12), and these help sooo much! i usually try to do them myself and then see if i get it right, but you have really helped me grasp the concepts, THANKS!
Good luck! Hope you crush those exams!
@kklola1 Yes, exactly! The physics in Angry Birds is pretty good. I think they calculate friction and momentum, too - just guessing based on how it all looks.
you sir are amazing.
hey teacher, how close does my catapult/cannon need to be to clear the wall (degrees) (dist) please,thanks, knew you would take the test
wht a nice lecturer sir thanx u 2 much free f cost
So I used [[[[[[[ v_f = v_0 + at ]]]]]] and I got t = 1.586 s, then I plugged it into
[[[[[[ y_f = y_0 + v_0*t + ((1/2)at^2) ]]]]] and I got 12.34 m, which means the rock still did
not go over the wall, but how to know which equation to use. x_f=vt or v_f = v_0 +at
Please can someone do this question as an example...(a car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal.The negligent driver leaves the car in neutral,and the emergency brakes are defective.The car rolls from rest down the incline with a constant acceleration of 4 meter per squared seconds for a distance of 50m to the edge of the cliff,which is 30m above the ocean. )......(a) find the length of time the car is in air.(b) find the car's position relative to the base of the cliff when the car lands in the ocean..
There are different ways to set up these problems. Your teacher may be doing it differently, but is certainly using a method that works out to be mathematically equivalent.
The way I set these up is this: The x component of the velocity is found by multiplying v0 by the cosine of the angle; the y component of the velocity is found by multiplying v0 by the sine of the angle.
for anyone who still has that question.. you don't use the peak height because the peak height doesn't tell you the height at the time you're to the castle wall, which is all that matters.
well if it cant make it over the wall at what angle do they need to aim to make it over the wall?
@derekowens This is a really stupid question, but what do you mean by the moment at the horizontal position of the wall? Do you mean when the trajectory ends?
Thanks for the practice, but why didn't the rock reach over the building? Is it because the distance if the rock is 4.1m ?
nice problem--well worked
والله فناااان
How did he find the acceleration in the vertical??
Thank you sir, GOD bless u
I used the height to get the time, and my range is over 45m, so does it mean that the rock can pass the wall depending on the solution?
I did what you did initially, that is solve for t=x/v, but then i realise that the title says:" does it go over the wall?" so i used the v(peak)=v(sub o at the y direction) + at to find out the time of flight first. But then it turns out to be wrong
Thanks a lot....!!!!
For the Horizontal component, why didn't you put the acceleration for it, but you did in the vertical component?
@Denseworldproduction Maybe. Tell me what info you are given, or just tell me the whole problem.
@derekowens its okay i got the answer, thanks for your help!
I have a question. Why did you use 2.89 s directly to solve for the height of the rock at that time? Wouldn't that be not correct? The rock will first go upward before going downward, so it will use different signs of gravitational acceleration. I'm confused.
i was thinking the same. He would have had to use the difference between the total time and the time it took to ascend. A projectile could also cross a certain point on its way down
Labyu mi²
hi just wondering,
why is it + 1/2 at^2? i thought the equation in the vertical direction was y = yo + voy t - (1/2 at^2) ????
very helpfull
I love how you explain
It helps me to understand more about how it works, thank you very much for the video❤❤❤
When I used trig to find the x and y components, I definitely did not get a full decimal number. I got 11sqrt(2)
Why is the acceleration negative? If it’s launched up shouldn’t the acceleration be positive?
can you show me how to do this problem if im given a height above ground as my start point, and angle but its asking the same question as above, how would i solve this? no velocity given
Thank you so much. I have the hardest time understanding most physics concepts and this has been incredibly helpful
@derekowens but i dont have velocity can you please explain in more detail
I think the answer is yes not negative.
Because, The rock would go maximum 49 meters horizontally. It would also go 41 meters. So, it is actually that the rock would cross the wall.
The question, though, isn't simply about the maximum distance and the maximum height. The maximum distance may be more than 45, and the max height may be more than 20. The question though, is how high is it after it has traveled 45 meters horizontally. That is the question we need to answer to see if it will clear the wall.
Is my final answer also correct? Because, i set my Scientific Calculator into 2 decimal places.
44.97m - 40.93m = 4.04m or 4m
then how would you get the total time traveled if its not 2.89s?
u didnt start from a height above the ground. lets say i started 5 m above the ground. Would i add 5 to teh 4.04 ?
For anyone who has seen the video recently.. For the distance, while calculating the time, he used 45 meters. That's the distance from the catapult to the wall, is it right just assume that's also the distance the rock travels (what he did in the video) although we haven't been given that? I started with the vertical component and used the Vf=Vi+at and got the time=1.6 seconds instead of what he got (2.89). Am I making a mistake here ?
Ahmad,
1.6 seconds is the correct value for the time to the peak, the highest point. That doesn't quite help us solve the problem, though. Using a time of 1.6 seconds we can find the height at the peak, and it turns out to be over 20 meters, but after the peak it starts coming back down, and we still don't know if it is above 20m at the time it reaches the wall.
So there are two other ways we can solve it. We can find the time it takes to go horizontally 45 meters, and then see if the height is above 20m at that time. Or we can find the time at which it is 20m high, and then see if it is horizontally beyond 45 meters at that time. Either way will work, but starting by finding the time to go horizontally is a bit easier in this particular problem. Hope that helps!
not to be a super geek but if you keep the exact values through out the problem, you will get a little less than 4m like around 3.99m. I know it's irrelevant here but I think it can make a huge difference in real world no? like .2m can let the rock pass over the wall if wall was hmm lets say 3.9m tall.... However, great videossssss....i learned a lot from them as you can see :P
For rock to climb the wall the speed should be nearly above 28m/s
Ilysm
Catapults having to do with rotational motion
Just when i thought that I have no hope in passing college physics.. then I encountered these amazing videos.. Thank you sir :))
big help for playing angry birds!
wow, I love you, totally helpin me out here, my teach explains horribly compared to you buddy! thanks !
I understand the equation for the Yf would be :
Yf=Yo+Vyit - 1/2gt^2 you added 1/2 gt^2 instead of subtracting it. Why
u saved my ass im forever indebted
mommy? sorry. mommy? sorry. mommy.
@@yansie smells like fruit
5:56. you lost me there.. doesn't the the total time of the flight equals to the total time it took for the distance to go horizontally? @.@
help!
horizontally
On the cosine and sine part, I was just wondering how you got 15.56? When I plug it in my calculator, I only got 11.55 for both.
Make sure your calculator is in degrees mode
is my method correct? since the wall is 20 meters high, i just solve for the maximum height of the projectile which is 12.33 meters and just by seeing the maximum height I can assume that it will not pass through the wall.
Yes, that is valid, at least in *this* case. If the maximum height were higher than 20 meters, you would have to do a little more work. Specifically, you would need to determine if it is above 20m at the same moment it is at the horizontal position of the wall. Your approach shows a good intuition for solving this quickly. The more general solution is also valuable, though.
oh and wots your teaching and example
I still don't get why the acceleration is negative.
The acceleration is *down* simply because gravity pulls down. For any problem, you can decide if you want up to be the positive direction or down to be the positive direction. Either way is okay, just decide and stick to it through the end of the problem. If we make up the positive direction, then down is the negative direction, and since gravity pulls down the acceleration is therefore negative. Hope that helps!
what if my question doesn’t give the initial velocity?
In that case, you should be given some other information that will either allow you to find it, or will allow you to find something that will allow you to find it. In other words, from what you are given, you can typically find *something*, then something else, then something else, until you find the particular value you are aiming to find.
I could not get it. Why the vertical time equals to the horizontal time check it out at 5:40 ???? I think we have to divide by 2 to find the time vertical. Please answer my question ,,,
I'll try to help. From one point along to the trajectory to a second point along the trajectory will take a certain time. We can consider the horizontal motion during this time or the vertical motion during this time, but the time should be the same in each case. There are problems where we need to consider half of the flight versus the whole flight, and in those problems we often end up multiplying or dividing by 2, but this problem is different. Here, we are finding the time for the rock to reach the wall, which is generally not going to be at the midpoint of the flight.
Thanks a lot, that's a very helpful answer
a00000z100 Hello :)
You get the time of the rock that traveled 45m away.
Hope this helps you.
in vertical
vot + 1/2at^2
where is the value of T?? Plz help me out
There is a difference between "T" and "t". Replace your value that you got for time with "t" in the equation. [ Vo(t) + 1/2 (a)(t)^2].
now we know how to attack the pigs....ITS ON! xD thanks tho idk how but ok ._.
how you get 40.9 m?
The rock needs to past 20m but the final answer that we get is 4.1m, so the answer is no
I AM REAlly confused.
R should be 45m. or higher
R = Vo²sin2(¢)/g
= (22)²sin(2×45)/(9.81)
= 49.34 m.
H should be 20m. or higher
H = Vo²sin²(¢)/2g
= (22)²sin²(45)/2(9.81)
= 12.33 m.
yes, it will reach??? 😂