A red ball is thrown up off the edge of a building 10 m tall with an initial speed of 4 m/s.

A) how long does it take for the red ball to strike the ground below?

B) If a black ball is dropped off the edge of the same building, how many seconds after the release of the red ball must the black ball be released so that they strike the ground at the same time

1 answer

A) Solve this equation for the time t:
Y = 10 + 4t -4.9 t^2 = 0
Take the positive root. There will be two solutions.

t = [-4 -sqrt(16 + 196)]/-9.8
= 1.89 s

B)The black ball that is dropped requires t = sqrt(2H/g)= 1.43 s to fall

To hit the ground at the same time as the red ball, delay its release by 1.89-1.43 = 0.46 seconds