Question bank/JEE Main/Physics

If the projectile hits the inclined plane perpendicularly when
thrown horizontally with $v_{0}$, from a tower of height H as shown then
the value of $\frac{4gH}{{(v_{0})}^{2}}$ is 2k. Find k.

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$a_{y} = gcos30^{\circ}$
$u_{x} = V_{0}cos30^{\circ}$
$u_{y} = V_{0}sin30^{\circ}$
$V_{x} = u_{x} + a_{x}t$
$0 = \frac{V_{0}\sqrt{3}}{2} - \frac{g}{2}t$
$t = \frac{v_{0}\sqrt{3}}{g}$....(i)
$S_{y} = u_{y}t + a_{y}t^{2}$
$(Hcos30^{\circ} - 0) = V_{0}sin30^{\circ}t + \frac{1}{2}gcos30^{\circ}t^{2}$....(ii)
From (i) (ii)
$\Rightarrow \frac{4gH}{{(v_{0})}^{2}} = 10$

Source: Grass-G / P-16 Physics paper 25 Nov First.docx