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The surface of a hill inclined at an angle 30° to the
horizontal. A stone is thrown from the summit of the hill (point A) at an initial speed 10 m/s at angle 60° to the
vertical. If the stone strikes the hill at point B as shown in
the figure, the distance between A and B is
(Take g = 10 m/s2)
Options:
horizontal. A stone is thrown from the summit of the hill (point A) at an initial speed 10 m/s at angle 60° to the
vertical. If the stone strikes the hill at point B as shown in
the figure, the distance between A and B is
(Take g = 10 m/s2)
Solution:
2095 Upvotes
Verified Answer
The correct answer is:
$20 \mathrm{~m}$
Component of velocity along incline $=\left(10 \cos 60^{\circ}\right)=5 \mathrm{~m} / \mathrm{s}$ and, $\mathrm{T}=\frac{2\left(10 \sin 60^{\circ}\right)}{\mathrm{g} \cos 30^{\circ}}=2 \mathrm{sec} \quad\left[\because \mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g} \cos \alpha}\right]$
$$
\begin{aligned}
& \text { So, } \mathrm{AB}=5 \times 2+\frac{1}{2} \times \mathrm{g} \sin 30^{\circ} \times 2^2 \\
& =10+10=20 \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
& \text { So, } \mathrm{AB}=5 \times 2+\frac{1}{2} \times \mathrm{g} \sin 30^{\circ} \times 2^2 \\
& =10+10=20 \mathrm{~m}
\end{aligned}
$$
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