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#1
This is the link to the question:
https://imgur.com/a/PBQ0exx
I'm not really sure how to do this. Any help is appreciated.
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1 month ago
#2
(Original post by CtrlAltDelirious)
This is the link to the question:
https://imgur.com/a/PBQ0exx
I'm not really sure how to do this. Any help is appreciated.
This is a projectile motion question. Think about the motion from the moment the ball is thrown horizontally until the moment it bounces on the ground - because of symmetry (and the fact that you are told there will be no energy loses when the ball bounces or due to air resistance), if the ball bounces at the point half way between them, the lowest point in the half-pipe, it will go on to reach the other person.

In this motion, you know this vertical information: how far the ball has to fall, how fast it is moving initially, and the acceleration. Horizontally, you know the acceleration, and with a little bit of work you can work out how far the ball has to travel (a sketch might help). So if you can work out the time taken using the vertical information, you can add this to the horizontal information and then go on to work out the initial horizontal speed that you want.
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#3
(Original post by Pangol)
This is a projectile motion question. Think about the motion from the moment the ball is thrown horizontally until the moment it bounces on the ground - because of symmetry (and the fact that you are told there will be no energy loses when the ball bounces or due to air resistance), if the ball bounces at the point half way between them, the lowest point in the half-pipe, it will go on to reach the other person.

In this motion, you know this vertical information: how far the ball has to fall, how fast it is moving initially, and the acceleration. Horizontally, you know the acceleration, and with a little bit of work you can work out how far the ball has to travel (a sketch might help). So if you can work out the time taken using the vertical information, you can add this to the horizontal information and then go on to work out the initial horizontal speed that you want.
I tried that already and this is how far I got:
https://imgur.com/a/3HxDRUZ
I'm not sure if I'm going wrong here, but do I need to solve for t as a quadratic at the bottom? For some reason it didn't seem right to me.
Last edited by CtrlAltDelirious; 1 month ago
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1 month ago
#4
(Original post by CtrlAltDelirious)
I tried that already and this is how far I got:
https://imgur.com/a/3HxDRUZ
I'm not sure if I'm going wrong here, but do I need to solve for t as a quadratic at the bottom? For some reason it didn't seem right to me.
That's not my interpretation of the question. I think you want a diagram like this.

I think I'm right because it makes the horizontal distance the ball has to travel come out nicely.
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#5
(Original post by Pangol)
That's not my interpretation of the question. I think you want a diagram like this.

I think I'm right because it makes the horizontal distance the ball has to travel come out nicely.
Thanks, that makes it a lot clearer. I've got time as 0.9035s (4s.f.), but how would I know the horizontal distance?
0
1 month ago
#6
(Original post by CtrlAltDelirious)
Thanks, that makes it a lot clearer. I've got time as 0.9035s (4s.f.), but how would I know the horizontal distance?
I agree with your value for t - might be more elegant to leave it in terms of g for now and do the actual calculation at the end.

For the horizontal distance, try drawing a triangle on the diagram. One side will be from the centre of the circle to the point where the ball is thrown.
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#7
(Original post by Pangol)
I agree with your value for t - might be more elegant to leave it in terms of g for now and do the actual calculation at the end.

For the horizontal distance, try drawing a triangle on the diagram. One side will be from the centre of the circle to the point where the ball is thrown.
I got that side to be 10m, but I can't see how to work out a hypotenuse or the other side.
0
1 month ago
#8
(Original post by CtrlAltDelirious)
Thanks, that makes it a lot clearer. I've got time as 0.9035s (4s.f.), but how would I know the horizontal distance?
I agree with your value for t - might be more elegant to leave it in terms of g for now and do the actual calculation at the end.

For the horizontal distance, try drawing a triangle on the diagram. One side will be from the centre of the circle to the point where the ball is thrown.

(Original post by CtrlAltDelirious)
I got that side to be 10m, but I can't see how to work out a hypotenuse or the other side.
Try using that side as the hypotenuse. Horizontal side is the length you want, vertical side should be easy to work out.
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