A particle of mass m is projected with a speed u from the ground at an angle θ = π 3 w.r.t.…

1. $\frac{3\sqrt{3}}{8}\frac{u^2}{g}$
2. $\frac{3\sqrt{2}}{4}\frac{u^2}{g}$
3. $\frac{5}{8}\frac{u^2}{g}$
4. $2\sqrt{2}\frac{u^2}{g}$

Solution

$p_i=p_f$
$mu+mucos\theta =2mv$
$\Rightarrow v=\frac{u\left(1+\cos {60}^0\right)}{2}=\frac{3}{4}u$
So horizontal range after collision $=vt$
$=v\sqrt{\frac{{2H}_{max}}{g}}$
$=\frac{3}{4}u\sqrt{\frac{2u^2{\sin }^2\left({60}^0\right)}{2g^2}}$
$=\frac{3}{4}{u}^2\frac{\sqrt{\frac{3}{4}}}{g}=\frac{3\sqrt{3}{u}^2}{8g}$

Asked in: JEE Main 2020 (09 Jan Shift 2)

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