Question

A stone is dropped from top of a tower of height 'h'. after 1 second another stone is dropped from the balcony 20m below the top . Both each the bottom at the same time instant . what is the value of h. ( g= 10m/s )

Answer 1

Let t be the time taken for the first ball to reach the ground.

Since the 2nd ball starts 1s later and the balls reach the ground simultaneously, we can say that the 2nd ball is in the air for (t-1)s.

We know from the equation of motion that

S=ut+0.5gt²

Here s is displacement and u is initial velocity.

Initial velocity for the both the balls is 0 as they are dropped from rest.

Let height of the tower be H.

Therefore for the first ball,

H=0.5gt²

And for the second ball

(H-25)=0.5g(t-1)²

Solving the equations, we get t=3s.

Substituting t in the H=0.5gt², we get H=45m

Here I have taken g=10m/s