A body of mass $M$ thrown horizontally with velocity $v$ from the top of the tower of height $H$ touches the…
Solution

$\begin{aligned} & 100=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} ; \quad \mathrm{x}=\frac{v}{2} \sqrt{\frac{2(4 \mathrm{H})}{\mathrm{g}}}=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} \\ & \Rightarrow \mathrm{x}=100\end{aligned}$
Asked in: JEE Main 2024 (08 Apr Shift 2)
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