Question

A ball is dropped from the top of a h metre high building it takes t seconds to reach the ground at time taken in travelling its first half height is​

Answer 1

Answer:

Distance travelled = h/2

Acceleration = 10 m/s^2

We know that:

s = ut + 1/2at^2

=> h/2 = 0(t) + 1/2(10)(t^2)

=> h = 10t^2

=> t^2 = h/10

=> $\tt{t = \sqrt{\frac{h}{10}}}$ s

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