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A body falling from rest under gravity passes a certain point $P$. It was at a distance of $400 \mathrm{~m}$ from $P, 4 \mathrm{~s}$ prior to passing through $P$. If $g=10 \mathrm{~m} / \mathrm{s}^2$, then the height above the point $\mathrm{P}$ from where the body began to fall is
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The correct answer is:
720 m
720 m
We have $h=\frac{1}{2}{g t^2}^2$ and $h+400=\frac{1}{2} g(t+4)^2$.
Subtracting we get $400=8 \mathrm{~g}+4 \mathrm{gt}$
$$
\begin{aligned}
& \Rightarrow \mathrm{t}=8 \mathrm{sec} \\
& \therefore \mathrm{h}=\frac{1}{2} \times 10 \times 64=320 \mathrm{~m} \\
& \therefore \text { Desired height }=320+400=720 \mathrm{~m} .
\end{aligned}
$$
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