A particle is projected vertically upwards. It will be at ^th of its greatest height at time and . . Then the ratio is

JEE MainPhysicsModerateMotion in 1D

A particle is projected vertically upwards. It will be at
$\left( \frac{3}{4} \right)$ ^th of its greatest height at time
$t = t_{1}$ and $t = t_{2}$ . $\left( t_{1} < t_{2} \right)$ . Then the
ratio $\left( t_{1}:t_{2} \right)$ is

Show answer and solution

H = \frac{u^{2}}{2g}

;

\frac{3}{4}\left( \frac{u^{2}}{2g} \right) = ut - \frac{1}{2}gt^{2} \Rightarrow t = \frac{u}{g} \pm \frac{u}{2g}

\Rightarrow t_{1} = \frac{u}{2g}

and

t_{2} = \frac{3u}{2g} \Rightarrow t_{1}:t_{2} = 1:3

Source: Grass-G / P-05 Physics paper 10 October.docx

Scan any doubt from your notes and get a whiteboard explanation.