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I'm doing a physics course in my spare time and the problem set is:

Ball A is dropped from rest from a building of height H exactly as ball B is thrown up vertically from the ground. When they collide A has twice the speed of B. If the collision occurs at height h, what is h/H?

It also gives a hint: Write equations for heights $y_A, y_B$ and velocities $v_A, v_B$. What can you say about them at the time of the collision?

So, I'm stuck at the beginning. I'm trying to find a displacement function for ball A, and I realized that I need to find a velocity function first, and the only thing I know is that the ball is moving under gravity so then $a_A(t) = -g$ and then integrated this means $v_A(t) = -1/2gt^2$.

So far so good, except that now I got confused. Should I integrate the velocity function? It doesn't feel like I should, and even so I'm not really sure how - so I looked at the problem solution.

The problem solution says the displacement function is $y_A(t) = H - 1/2gt^2$ - which looks a lot like my velocity function! I am confused. The gravitational force accelerates mass ($F = ma$ amirite), so how can this be?

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2 Answers

up vote 2 down vote accepted

You double integrated to get the -(1/2)gt^2 !

Integrate(-g) = -gt + C (C=0 for starting at rest)

Integrate(-gt) = -(1/2)gt^2 + C (C=H for starting at top of building)

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I see my fault: the integral of $a_A(t) = -g$ is $v_A(t) = -gt$, and therefore the displacement function is $y_A(t) = H - 1/2gt^2$.

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