dd1234
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Just need some help with two questions:

1) Bill is standing on the flat roof of a house 6m high and Ben is on the ground vertically below. Ben throws a ball up to Bill with a speed of 16.7 m/s and Bill catches it on the way down. How long is the ball in the air and how fast is it moving when Bill catches it?

I have worked out that the speed when Bill catches it is 12.7 m/s which is correct, but am having trouble finding the time, here are the values I am using to find the time:

S= 6
U=16.7
A=-9.8
T=?

The second question is a graph question, having trouble with both a and b,

Any help would be great, thanks!
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Pictraz
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For the first question, split up the motion of the ball into 2 stages. Stage 1 is the ball going up from Ben and becoming stationary. Stage 2 is the ball coming down from being stationary and getting caught by Bill. Add up the two times.

For the graphs question, remember that acceleration is constant, and the gradient of a speed-time graph is acceleration. So what does that mean about the shape of the graph? And what happens to a ball's velocity when it is thrown into the air and is just about to fall back down?
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Joseph.S
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(Original post by dd1234)

For the second question: is part A wrong because the line isn't straight and therefore doesn't show a constant velocity? For part b I would have thought that the third graph is correct because it shows the velocity when it is in the air is 0 at 4 seconds, why is this not correct?


Thanks for your help!
I think it is because we are talking about velocity, which includes direction. So it would continue under the graph until -u.

As it goes in the opposite direction. H

Hope I helped.
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dd1234
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(Original post by Pictraz)
For the first question, split up the motion of the ball into 2 stages. Stage 1 is the ball going up from Ben and becoming stationary. Stage 2 is the ball coming down from being stationary and getting caught by Bill. Add up the two times.

For the graphs question, remember that acceleration is constant, and the gradient of a speed-time graph is acceleration. So what does that mean about the shape of the graph? And what happens to a ball's velocity when it is thrown into the air and is just about to fall back down?
For the first question, I did get 3 seconds ( the correct answer), but only when I calculated the time for the first stage using:
S=6
U=16.7
A=-9.8
T=? which gave me a quadratic which I solved and got 3 and 0.408 - do I still have to calculate the second stage?

For the second question: is part A wrong because the line isn't straight and therefore doesn't show a constant velocity? For part b I would have thought that the third graph is correct because it shows the velocity when it is in the air is 0 at 4 seconds, why is this not correct?


Thanks for your help!
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Pictraz
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No problem.

For the first question, 3 is the correct answer because you are calculating the time for the ball to go up and come back down again in a single step. With a quadratic, you get two answers. This is because the ball passes the roof at 6m twice - once going up and once coming back down. The 0.408 is the time taken for it to reach that height going up, and 3 is the time taken coming down. Therefore 3 is the answer. This method is quicker but a bit more risky if you're not sure, but you've got the right answer, so that's good

I'd have thought it's C as well. Is it not?
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dd1234
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(Original post by Pictraz)
No problem.

For the first question, 3 is the correct answer because you are calculating the time for the ball to go up and come back down again in a single step. With a quadratic, you get two answers. This is because the ball passes the roof at 6m twice - once going up and once coming back down. The 0.408 is the time taken for it to reach that height going up, and 3 is the time taken coming down. Therefore 3 is the answer. This method is quicker but a bit more risky if you're not sure, but you've got the right answer, so that's good

I'd have thought it's C as well. Is it not?
Great thanks, if I were to do the other method, which values would I use for the stages?
It says that neither graph is correct, but I think its not c because it has a negative velocity due to a change in direction
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Pictraz
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(Original post by dd1234)
Great thanks, if I were to do the other method, which values would I use for the stages?
It says that neither graph is correct, but I think its not c because it has a negative velocity due to a change in direction
So you would set v equal to 0 for the first stage of it going up, and find the time and distance. Then you would do the same, but this time set u equal to 0 and find the values you need.


The graph question mentions speed specifically, so I'd have thought it was just a modular graph...

But if not, it would simply go into the negative side, becoming a straight line and intersecting x at 4s.
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dd1234
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(Original post by Pictraz)
So you would set v equal to 0 for the first stage of it going up, and find the time and distance. Then you would do the same, but this time set u equal to 0 and find the values you need.


The graph question mentions speed specifically, so I'd have thought it was just a modular graph...

But if not, it would simply go into the negative side, becoming a straight line and intersecting x at 4s.
Would I have to find the distance for the first stage or could I just calculate the time using the u,v and a values?
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brianeverit
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(Original post by dd1234)
Would I have to find the distance for the first stage or could I just calculate the time using the u,v and a values?
Just use s=ut+\frac{1}{2}at^2\mathrm{\ with\ } s=6, u=16.7 \mathrm{\ and\ }a=-9.8 and solve the quadratic. The smaller answer is the time when the ball is going up and the larger the time when it is on the way down.
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dd1234
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(Original post by brianeverit)
Just use s=ut+\frac{1}{2}at^2\mathrm{\ with\ } s=6, u=16.7 \mathrm{\ and\ }a=-9.8 and solve the quadratic. The smaller answer is the time when the ball is going up and the larger the time when it is on the way down.
So would I have to add them both together to find the time it is in the air for?
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brianeverit
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(Original post by dd1234)
So would I have to add them both together to find the time it is in the air for?
No. The larger of the two answers is the time for which it is in the air
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dd1234
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(Original post by brianeverit)
No. The larger of the two answers is the time for which it is in the air
Why wouldn't we add them together if the smaller one is the time for it going down and the larger of the time it is going up, isn't it in the air this whole time?

Thanks
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brianeverit
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(Original post by dd1234)
Why wouldn't we add them together if the smaller one is the time for it going down and the larger of the time it is going up, isn't it in the air this whole time?

Thanks
The larger answer is the time from projection not the time from when it passes the point on the way up.
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dd1234
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(Original post by brianeverit)
The larger answer is the time from projection not the time from when it passes the point on the way up.
Okay, I understand now,
thanks a lot!
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brianeverit
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(Original post by dd1234)
Okay, I understand now,
thanks a lot!
You're welcome
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