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A boy playing on the roof of a $10 \mathrm{~m}$ high building throws a ball with a speed of $10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of $10 \mathrm{~m}$ from the ground?
$$
\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]
$$
Options:
$$
\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\right]
$$
Solution:
1816 Upvotes
Verified Answer
The correct answer is:
$8.66 \mathrm{~m}$
$8.66 \mathrm{~m}$
From, the question if the horizontal distance is none other than the horizontal range on the level of the roof of building
$$
\text { Range }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{(10)^2 \sin (2 \times 30)}{\mathrm{g}}=\frac{10 \times 10 \times \sqrt{3}}{2 \times 10}=8.66
$$
$$
\text { Range }=\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}=\frac{(10)^2 \sin (2 \times 30)}{\mathrm{g}}=\frac{10 \times 10 \times \sqrt{3}}{2 \times 10}=8.66
$$
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