Question

A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after time. (A)         v2 2gh 1 1 g v (B)         v2 2gh 1 1 g v (C) 1/ 2 v2 2gh 1 g v      (D) 1/ 2 v2 2gh 1 g v     

Answer 1

Hey Dear,

● Answer -

T = v/g [1 + √(1+2gh/v²)]

◆ Explanation -

While reaching the maximum height of the ball can be,

v' = v + at

0 = v - gt

t = v/g

Time taken to come down upto tower is same as for going up. i.e.

t = v/g

Time taken by the ball to reach ground from height h -

h = vt' + ½ at'²

at'² + 2vt' - 2h = 0

Solving this quadratic eqn,

t' = (v/g) √(1 + 2gh/v²)

Therefore, total time required to reach ground will be -

T = v/g + v/g √(1+2gh/v²)

T = v/g [1 + √(1 + 2gh/v²)]

Thanks for asking...