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A body is projected with a speed $u \mathrm{~m} / \mathrm{s}$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3 / 4$ th of the initial kinetic energy. The value of $\beta$ is :
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Verified Answer
The correct answer is:
$30^{\circ}$
Hints: (K.E.) at maximum height $=\frac{1}{2} m\left(u^2 \cos ^2 \beta\right)$
$\mathrm{K} . \mathrm{E} .=\mathrm{K} \cos ^2 \beta$
Here, $K \cos ^2 \beta=\frac{3}{4} \mathrm{~K}$
$$
\begin{aligned}
& \cos \beta=\frac{\sqrt{3}}{2} \\
& \beta=30^{\circ}
\end{aligned}
$$
$\mathrm{K} . \mathrm{E} .=\mathrm{K} \cos ^2 \beta$
Here, $K \cos ^2 \beta=\frac{3}{4} \mathrm{~K}$
$$
\begin{aligned}
& \cos \beta=\frac{\sqrt{3}}{2} \\
& \beta=30^{\circ}
\end{aligned}
$$
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