A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is the same as the time taken to come down.

During vertical upward motion of the stone

Given that S=h, u=u, v=0 and a= -g

And if ‘t’ is the time taken by the ball to reach the height ‘h’, then by applying a second equation of motion, you get – 2gh = v²-u².

Hence, u= √2gh

Applying the first equation of motion you get v = u-gt

Therefore, t = u/g= (√2h/g)

Now, for vertical downward motion of the stone

S=h, u=u and a= g

Given that “v” is the velocity of the ball with which it hits the ground and ‘t” is the time taken for the ball to reach the ground.

Here, applying second motion of the equation gives you

2gh = v²

So, v = √2gh

Now, applying second motion of the equation

v= u + gt

Hence, t= v/g = (√2h/g)

This proves that the time taken by the stone to go up is the same as the time taken to come down.