Question

From the top of a tower, a stone was projected vertically upwards with a velocity v. When the stone was at a distance h below the top of the tower, it had a velocity that was three times what it was at a distance 'h' above the top of the tower. Find the maximum height attained by the stone.

Answer 1

Considering the top of the tower as an origin and all vectors pointing upwards positive.Let the velocity of stone at height +h be v and at height -h be -ve.

We have v² = u² -2gh

or,(-3v)² = u² -2g(-h)

or,9v² = u² -2gh

or 9(u² - 2gh) =u² +2gh

or, 9u² - 18gh = u² + 2gh

or, 8u² = 20gh

or, u² = 5/2 gh.

If the maximum height attained be H, then (0)² =u² - 2gh

or, 0 =5/2gh - 2gh

or, H = 5/4 h.