@@Rawan-cr8gqwell if you use the path that the boat takes, you will still get the same answer, however you will have to use V_boat,land so 517.6m/2.15m/s gives you approximately 240s Edit: if you are asking why taking the other path works, it's because 500m can be written as 517.6 (which is the path that the boat takes) multiplied by cos(15°) and V_B,R can be written as 2.15 (which is V_B,L) multiplied by cos(15°) So when you divide them, the two cos(15°)'s cancel and you are left with 517.6 divided by 2.15 (which, again, is like saying actual distance that the boat takes divided by V_B,L)
@@ezxd5192 Correct. The confusing part of all this is that the 7.5 km/h is NOT the velocity of the boat, but rather it is the NORTHWARD component of the boat's velocity (It's speed ACROSS the river). IF the boat could only travel 7.5 km/h through the water, then it would take longer because the distance travelled (the hypoteneuse) is longer than the adjacent 500 meters.
Hey man I have a problem ive asked three different people for help on and I can’t figure it out, it basically asks if you kicked a ball at an angle after a given amount of time has passed, and it has a certain horizontal distance, what was the initial velocity the ball, this ball then lands on a ledge, and you have to calculate its velocity when it lands on that ledge, and then the height of the ledge, basically a projectile launched onto a ledge problem. Great video btw!
Sorry to be so off topic but does anyone know a tool to log back into an instagram account?? I was stupid lost the password. I appreciate any help you can offer me!
@Arlo Ayaan i really appreciate your reply. I found the site on google and Im trying it out now. Takes a while so I will reply here later when my account password hopefully is recovered.
How would I solve a problem similar to problem 2 if I know the velocity of the boat in still water instead of the velocity of the boat relative to the river
the velocity of the boat in still water is the same as the velocity of the boat relative to the river, because the boat flows with the river. Therefore, it doesn't matter how fast the river is moving. Boat's velocity to the river will always be the same.
@@galenhan1719 if question has velocity in still water you would not do pythagorean thereom to figure out distance it travelled horizontally right? You would just do 500/velocity of still water and times that by the current?
one thing though, when 2'2 + 7.5'2 and placed in a square root, the answer is 7.762087348 which rounded off two decimal placed is 7.76 but not, 7.75 there were more point where you don't use the calculator value, will that be a problem if we use the calculator value?
thank you man i was wondering wtf to do these kind of questions, ive recently been doing yr 11 physics despite being 2 years underaged and i think i lack the thinking kind of thing so thank you for this video
i HATE THESE PROBLEMS going to watch the video hopefully i will walk away as an actual person with a working brain instead of whatever jumbled rotten spinach milkshake is fogging my abilities right now
Literal lifesaver! Your physics channel is the best!
You got this!
At 9:23 why did you choose 500m? Why didn't you choose the path of the boat after it gets pushed since that's the actual path it will take?
I have the same question, did you find the answer?
@@Rawan-cr8gqwell if you use the path that the boat takes, you will still get the same answer, however you will have to use V_boat,land so 517.6m/2.15m/s gives you approximately 240s
Edit: if you are asking why taking the other path works, it's because 500m can be written as 517.6 (which is the path that the boat takes) multiplied by cos(15°) and V_B,R can be written as 2.15 (which is V_B,L) multiplied by cos(15°)
So when you divide them, the two cos(15°)'s cancel and you are left with 517.6 divided by 2.15 (which, again, is like saying actual distance that the boat takes divided by V_B,L)
@@ezxd5192 Oh I got it now, Thank you
@@ezxd5192 Correct. The confusing part of all this is that the 7.5 km/h is NOT the velocity of the boat, but rather it is the NORTHWARD component of the boat's velocity (It's speed ACROSS the river). IF the boat could only travel 7.5 km/h through the water, then it would take longer because the distance travelled (the hypoteneuse) is longer than the adjacent 500 meters.
It's great, simple and easy to understand 👌. Thank you
It just help me solve my assignment question, thanks very much
You gotta believe me.... It really helped me understand rather than listening to my physics teacher and learning nothing😒
Thank you 🤍
Same here
Why is everyone hating their physics professors? 🤣. Mine is worst though.
in the third problem, can you just calculate the angle using sin?
This Helped Me ALOT Thank you!
Hey man I have a problem ive asked three different people for help on and I can’t figure it out, it basically asks if you kicked a ball at an angle after a given amount of time has passed, and it has a certain horizontal distance, what was the initial velocity the ball, this ball then lands on a ledge, and you have to calculate its velocity when it lands on that ledge, and then the height of the ledge, basically a projectile launched onto a ledge problem. Great video btw!
Sorry to be so off topic but does anyone know a tool to log back into an instagram account??
I was stupid lost the password. I appreciate any help you can offer me!
@Arlo Ayaan i really appreciate your reply. I found the site on google and Im trying it out now.
Takes a while so I will reply here later when my account password hopefully is recovered.
@Arlo Ayaan It worked and I finally got access to my account again. I am so happy!
Thank you so much, you saved my ass!
@Cristiano Kelvin No problem =)
@@arloayaan9893 Obvious advertising
How would I solve a problem similar to problem 2 if I know the velocity of the boat in still water instead of the velocity of the boat relative to the river
the velocity of the boat in still water is the same as the velocity of the boat relative to the river, because the boat flows with the river. Therefore, it doesn't matter how fast the river is moving. Boat's velocity to the river will always be the same.
@@galenhan1719 if question has velocity in still water you would not do pythagorean thereom to figure out distance it travelled horizontally right? You would just do 500/velocity of still water and times that by the current?
In problem 3 can we use sin^-1(2/7.5) we would get the same value without the need of using all these steps.
thats what i also did but you get 15 deg after rounding whilst he gets 16 deg due to rounding off set
The d=vt when the boat is going down the stream wouldn't be the velocity x hypotenuse?
one thing though, when 2'2 + 7.5'2 and placed in a square root, the answer is 7.762087348 which rounded off two decimal placed is 7.76 but not, 7.75
there were more point where you don't use the calculator value, will that be a problem if we use the calculator value?
I think this is a resultant velocity problem, not a relative velocity problem.
huge thanks
thank you king
Bro thank you❤❤
Thank you
thank you man i was wondering wtf to do these kind of questions, ive recently been doing yr 11 physics despite being 2 years underaged and i think i lack the thinking kind of thing so thank you for this video
I thought i saw Physics NAIJA😭😭😭.. i thought it was an African channel for a sec there
Nice one
In indian these problems were taught in 11 th standard
And much more tougher these came in JEE advance
In indian these problems were taught in 11 th standard
And much more tougher these came in JEE advance 3:04
Thank you very much sir.!
i HATE THESE PROBLEMS going to watch the video hopefully i will walk away as an actual person with a working brain instead of whatever jumbled rotten spinach milkshake is fogging my abilities right now
😂
lock in gang
@@attraxia✌🏿
LMFAOAOAOA
pretty good video
First problem.
Vb,r is actually the hypothenuse in the equation
Thank you!!!
Thank you!
güzeldi helal
THANKYOUUUUUU
love love it
I don't like these type of questions sir I hate them
In indian these problems were taught in 11 th standard
And much more tougher these came in JEE advance
In indian these problems were taught in 11 th standard
And much more tougher these came in JEE advance
In indian these problems were taught in 11 th standard
And much more tougher these came in JEE advance