Question

A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant a ball is dropped from rest from a building 14 m high. After how long will the balls be at the same height?

Answer 1

Сoordinate of the first ball depends on time as follows:
x1 = 19.2t - 9.8t^2/2
19.2 - initial speed
9.8 - gravitational acceleration
Сoordinate of the second ball depends on time as follows:
14 - 9.8t^2/2
14 - initial coordinate
When the balls will be at the same height their coordinates will be equal:
x1 = x2:
19.2t - 9.8t^2/2 = 14 - 9.8t^2/2
t = 14/19.2 = 0.73 seconds
Answer: 0.73 sec