A ball is projected from the ground at an angle of 45 o with the horizontal surface. It reaches a maximum…

Solution

$H_1=\frac{u^2{\sin }^2{45}^o}{2g}=120$
$\Rightarrow \frac{u^2}{4g}=120 ...\left(\mathrm{i}\right)$
when half of kinetic energy is lost $v=\frac{u}{\sqrt{2}}$
$H_2=\frac{{\left(\frac{u}{\sqrt{2}}\right)}^2{\sin }^2 30°}{2g}=\frac{u^2}{16g} ...\left(\mathrm{ii}\right)$
From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$,
$H_2=\frac{H_1}{4}=30 \mathrm{m}$

Asked in: JEE Advanced 2018 (Paper 2)

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