Find where one object surpasses another.

Question

You are standing on top of a 219 m tall building. A drone operated by your friend is headed straight down, at a speed of 28.0 m/s. Just as the drone passes you, you release a tennis ball off the roof. How fast is the ball going when it passes the drone?

I've tried to make one equation equal another but it keeps marking it as wrong, so i'm not sure where i'm making a mistake

Answer 1

The drone is going down with constant velocity of 28 m/s. We start the time as drone passes you. Assume that after time t drone has covered a vertical distance S and it is given as

S=u*t =28t ....(1) (We used S=ut+1/2at^2 and put a=0 for drone)

Now assume that ball meets the drone at this vertical distance given above.

But for ball which is falling with acceleration due to gravity, the distance S is given by

S=u*t+1/2 g*t^2 =1/2 g*t^2 = 4.9t^2......(2) (ball starts from rest, so u=0 and a=g=9.8 m/s^2 )

Equating eqn (1) with (2), we get

28t= 4.9t^2

=> t=28/4.9= 40/7 secs

This is the time when ball passes the drone.

Now we use equation

v=u+gt =gt = 9.8t ........(3)

for ball to find the velocity of the ball at time when it passes the drone.

Putting t=40/7 in equation (3)

v= 9.8*40/7= 1.4*40=56 m/s

This is required velocity of ball when it passes the drone.