Question

A stone is thrown from the top of a building with an initial velocity at 20m/s down ward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Answer 1

Answer:

given ,

u = 20 m/s

a = g = 10 m/s^2

S = H = 60 m

we will now find the final velocity on touching the ground.

v^2 = u^2 + 2gH

v^2 = 400 + 1200

      = 1600

v = √1600

  = 40 m/s

by using the first equation of motion ,

t = v-u / g

 = 40 - 20 / 10

 = 20 / 10

 = 2 sec

Hence , the body will take 2 seconds to reach the ground.

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