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A ball is thrown at $30^{\circ}$ with horizontal, from the top of roof $20 \mathrm{~m}$ high with a speed of $13 \mathrm{~ms}^{-1}$. At what distance from the throwing point will the ball, once again be at a height of $20 \mathrm{~m}$ from the ground? $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$14.6 \mathrm{~m}$
The situation is as shown below.
Given, $\theta=30^{\circ}$, initial speed, $u=13 \mathrm{~ms}^{-1}$
$$
\begin{aligned}
\text { Range } & =\frac{u^2 \sin 2 \theta}{g} \\
& =\frac{(13)^2 \times \sin 60^{\circ}}{10}=\frac{169 \times \sqrt{3}}{10 \times 2}=14.6 \mathrm{~m}
\end{aligned}
$$
$\therefore$ Range, $R=14.6 \mathrm{~m}$
Hence, the ball once again be at a height of $20 \mathrm{~m}$ from the ground at $14.6 \mathrm{~m}$ from the point of projection.
Given, $\theta=30^{\circ}$, initial speed, $u=13 \mathrm{~ms}^{-1}$
$$
\begin{aligned}
\text { Range } & =\frac{u^2 \sin 2 \theta}{g} \\
& =\frac{(13)^2 \times \sin 60^{\circ}}{10}=\frac{169 \times \sqrt{3}}{10 \times 2}=14.6 \mathrm{~m}
\end{aligned}
$$
$\therefore$ Range, $R=14.6 \mathrm{~m}$
Hence, the ball once again be at a height of $20 \mathrm{~m}$ from the ground at $14.6 \mathrm{~m}$ from the point of projection.
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