Question

stone dropped in a well hits the water surface after 2 sec . what is the depth of well and with what speed will the stone hit the water surface

Answer 1

u = 0
t = 2sec
a = -g = 9.8m/s^2

S = ut + 1/2at^2
S = 0 + (-19.6)
S = -19.6m

v = u + at
v = -19.6 m/s

as body is falling from a height so , depth of the well will be 19.6m or 20m

and speed with which it strikes the surface of water = 19.6m/s or 20m/s

Hope it helps....

Answer 2

Answer-- s=ut+1/2at²(2nd equation of motion)

Here s=height or depth of the well

S=0×2+1/2×9.8(gravity)×2²

=0+ 9.8×2

=19.6m

It is the depth of well

Now,

As acceleration is happening as gravity

So,

V=u+at(1st equation of motion)

Here V is the speed at which it hits water surface

V=0+9.8(gravity)×2

=19.6 m/s

Note:u is 0 as first the stone was in hand with zero velocity

Note:Dont confuse that how depth and velocity is same as acceleration is happening and that is the speed at which it hits and not the average speed.

Here gravity is positive as the ball is going downwards