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

answer :

explanation : a ball is thrown vertically upwards from the top of a tower of height h with velocity v.

so, Let's time taken to reach maximum height H, where velocity of ball becomes zero.

using formula, v = u + at

here v = 0, u = v , a = -g

and t = v/g .....(1)

and height from the top of tower , H

using formula, v² = u² + 2as

here, v = 0, u = v, a = -g and s = -H

then, H = v²/2g .....(2)

after reaching maximum height, ball starts to fall downward due to gravitational force.

time taken by ball to fall the ground , T

using formula , s = ut + 1/2 at²

here s = h + H, u = 0, a = g and t = T

so, h + H = 0 + 1/2 (g) T²

or, h + v²/2g = 1/2 gT²

or, 2gh + v² = (gT)²

or, T = √(v² + 2gh)/g

now, total time taken by ball = t + T

= v/g + √(v² + 2gh)/g

= {v + √(v² + 2gh)}/g

Answer 2

Answer:

hey mate your answer is

Explanation:

D