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The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
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Verified Answer
The correct answer is:
$60^{\circ}$
The speed of a projectile at its maximum height
$$
\begin{aligned}
\mathrm{v}^{\prime} & =\mathrm{v}_0 \cos \theta \\
\frac{\mathrm{v}_0}{2} & =\mathrm{v}_0 \cos \theta \\
\cos \theta & =\frac{1}{2} \\
\theta & =60^{\circ}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{v}^{\prime} & =\mathrm{v}_0 \cos \theta \\
\frac{\mathrm{v}_0}{2} & =\mathrm{v}_0 \cos \theta \\
\cos \theta & =\frac{1}{2} \\
\theta & =60^{\circ}
\end{aligned}
$$
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