Question

A body projected vertically up cross points. A and B separated by 28 m with velocities one-third and one-fourth of the initial velocity, respectively. What is the maximum height reached by it above the ground?​

Answer 1

Let initial velocity of the body = u

Velocity of the body at point A =u/3

Velocity of the body at point B = u/4

Apply v^2 −u^2 = 2aS for points A and B:

( u/4)^2−( u/3)^2=2(−g)×28

On solving we get: u^2=1152g

Now apply direct formula for maximum height

Hmax=u^2/2g

H = 1152g/2g

= 576 m