Question

A body of mass m thrown vertically upwards attains a maximum height h. at what height will it's kinetic energy by 75% of its initial value

Answer 1

Let vf = final velocity

Vi = initial velocity.

Acc to ques

K.Efinal=3/4 K.Einitial (75% of initial)

1/2mvf^2 = 3/4(1/2)mvi^2

vf^2 = (3/4)vi^2…….. eqn 1

Now if H is the maximum height achieved by the ball when vf=0 therefore,

vf^2 = vi^2 −2gs

0=vi^2 − 2gH

vi^2 = 2gH ………… eqn 2

Also for the height h when the K.E. is 75% it's maximum, we use the kinematic equation using vf obtained earlier in eqn 1

(3/4)vi^2 = vi^2 −2gh

2gh=(1/4)vi^2

Putting equation 2 here -

2gh = (1/4)2gH

h=H/4.