Question

17. A stone is dropped into a well of 20 m deep.
Another stone is thrown downward with
velocity 'v' one second later. If both stones
reach the water surface in the well simultan-
eously, v is equal to (g = 10 ms 2)​

Answer 1

Explanation:

Time taken by the first stone to reach the water surface would be ,

s=ut+1/2gt^2

now since, u=0m/s [for the first stone]

20=1/2x10xT^2

Therefore, T comes out to be 2 seconds .

Now since the other stone is dropped 1 second later,

the time taken by the stone[which is thrown with a velocity v]would be ,

20= [Vx[T-1]] +1/2g[T-1]^2

20=v+5...........[since T-1=1]

v=15m/s .