Question

A body projected vertically up crosses point 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 X represent initial speed. Then speed at point A is X/3 and speed at B is X/4.

Distance between A and B is equal to:

$Dab = ((X/3)^2-(X/4)^2)/(2*g) = 28$

Solving the equation for X, obtaining X = 106.3 m/s.

Maximum height is determined as follows.

$Hmax=X^2/(2*g) = 106.3^2/(2*9.81) = 576 m$