Question

a body is released from the top of a tower of height H metres after 2 second it is stopped and then instantaneously released what will be its height its height after next 2 seconds

Answer 1

The INITIAL VELOCITY of the body be : u

      FINAL VELOCITY be : v ; which will be ZERO '0'

      v= u +  at

=>   0=u + 10 * 2  ( v=0 as the body has been stopped once during its passage to ground )

=> u = -20 m/s (aprox.)

To find out the height ;

                                              v^2= u^2 - 2as

                                                 0 = (-20)^2 - 2 * 10 * s

                                    -400/-20 = s

=>                                             s = 20 m

Thus the HEIGHT be 20 mts

Answer 2

Answer:

Explanation:

Let H be the height of the tower

Take g=10m/s

u=0

After 2sec

h=1/2gt^²

=1/2*10*2*2

=20 m

Again after 2 sec

h=20m

=The height travelled by the body after 4 sec is H-40 m