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.
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Expert's answer
2013-08-20T08:36:19-0400
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.
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