Question

A stone is dropped from the top of the tower of height 200m. Simultaneously, another stone is thrown vertically upwards from the foot of the tower with a velocity 50m/s. Find when and where the stones meet.

Answer 1

The equation of the motion for the first stone:
x1 = 200 - gt^2 / 2
The equation of the motion for the second stone:
x2 = 50t - gt^2 / 2
The stones meet when x1 = x2:
200 - gt^2 / 2 = 50t - gt^2 / 2
200 = 50t
t = 4
x1 = x2 = 50*4 - g*16/2 = 200 - 8*9.8 = 121.6
So the stones meet after 4 seconds at height 121.6 m.