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