Question

The stone is released from the top of a tower of height 5m. Consider the acceleration produced in the stone is 10 m/s2 . The velocity of the stone just before touching the ground will be......

Answer 1

Answer:

initial velocity (u) =0m/sec

acceleration (a) = 10m/sec

distance (s) =5m

now, using the iiird eq. of motion

I. e, 2as=v^2-u^2

2*10*5=v^2-0^2

100=v^2-0

under root 100=v

10m/sec=v

may it will help u