# Answer to Question #34303 in Mechanics | Relativity for Raison joseph

Question #34303

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.

Expert's answer

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.

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.

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