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A stone is dropped from the top of a tower of height h=60m. Simultaneously another stone is projected vertically upwards from the foot of the tower. They meet at a height $\frac{2 h}{3}$ from the ground level. The initial velocity(ms^–1) of the stone projected upwards is (g=10ms^–2)

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Detailed Solution

Time taken by upper stone to cover h/3 = 20 m is 2 sec

Time by lower stone to cover 40 m = 2 sec

Finding initial velocity

H = ut - 1/2 gt²

40 = u(2) - 1/2(10)(2)²

u = 30 m/s

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