Ball A is thrown upward from the top of a building and takes 12 s to hit the ground. Ball B is thrown

downward with the same speed as Ball A and takes 7 s to hit the ground. How long does it take for ball A to reach it
peak?

Ball A

Hi and Vi
v = Vi - g t
h = Hi + Vi t - 4.9 t^2
when t = 12 , h = 0
0 = Hi + 12 Vi -4.9 (144)
Hi + 12 Vi = 706

Ball B
Vi' = -Vi
Hi = Hi
v = -Vi - g t
0 = Hi - Vi (7) - 4.9 (49)
Hi - 7 Vi = 240

subtract the equations to eliminate Hi
19 Vi = 466
Vi = 24.5 m/s

Now a new problem - we launch a ball up with initial speed 24.5 m/s
how long until v = 0?
v = Vi - gt
0 = 24.5 - 9.81 t
t = 2.5 seconds

Plug and chug, plug and chug :)

1 s

-0.025

To find out how long it takes for Ball A to reach its peak, let's break down the problem and analyze the given information.

We know that Ball A takes a total of 12 seconds to hit the ground. From this information, we can determine the total time it takes for Ball A to go up and then come back down. Since the time taken to reach the peak is expected to be half of the total time, we can use this information to find the time it takes for Ball A to reach its peak.

1. Determining the total time for Ball A:
Since it took Ball A a total of 12 seconds to hit the ground, we know that it took the same amount of time to go up and come back down. Therefore, Ball A took 12/2 = 6 seconds to reach its peak.

Hence, Ball A takes 6 seconds to reach its peak.

Please note that this solution assumes no air resistance and a constant gravitational force acting on both balls.