a ball is released from the top of a 67.0 m high building. At the same instant a second ball is shot straight up from the ground with an initial speed of 20.0 m/s, directly below the point where the first ball was dropped. Ignore air drag and calculate

Question

a. the height above the ground where the balls collide.
b. the velocity (magnitude and direction) of each ball just before the collision.

Answer 1

dropped ball:
hf=67-4.9t^2
thrown ball:
hf=20t-4.9t^2
set them equal
20t-4.9t^2=67-4.9t^2
solve for collision time t.

hf use either equation to solve.

vf=vi-4.9t^2 find vf for both balls, using time t.

Answer 2

h1 = -1/2 g t^2 + 67.0

h2 = -1/2 g t^2 + 20.0 t

set the heights equal and solve for t
... substitute back to find the collision height

b. v1 = g t ... downward

v2 = 20.0 - gt ... downward

the 2nd ball reaches its max height and begins to fall back down before the 1st ball catches it