Question

A body of mass m thrown vertically upwards attains a maximum heights h. At what height will its kinetic energy by $75\%$ of its value ?

• A. $\frac{h}{6}$
• B. $\frac{h}{5}$
• C. $\frac{h}{4}$
• D. $\frac{h}{3}$

Answer 1

The correct option is C $\frac{h}{4}$

${K.E}_{final}=\frac{3}{4}{K.E}_{initial}$

above statement is the 75% of the initial kinetic energy

$\frac{1}{2}m{v}_f^2=\frac{3}{4}.\frac{1}{2}m{v}_i^2\Rightarrow v_f^2=\frac{3}{4}{v}_i^2⟶\left(1\right)$

now if h is the maximum height acheive by the ball when $v_f$=0 then using the kinematic equation give

${K.E}_{final}={K.E}_{initial}-{P.E}_{final}\Rightarrow \frac{1}{2}m{v}_f^2=\frac{1}{2}m{v}_i^2-mgh\Rightarrow v_f^2=v_i^2-2gh\Rightarrow 0=v_i^2-2gh\Rightarrow v_i^2=2gh⟶\left(2\right)$

Now for the height H when the K.E is 75%, we use the kinemtic equation using euation (1)

$\frac{3}{4}{v}_i^2=v_i^2-2gH\Rightarrow 2gH=\frac{1}{4}{v}_i^{2}puttingequation\left(2\right)here2gH=\frac{1}{4}2gh\therefore H=\frac{h}{4}$