Q. A boy playing on the roof of a 10 m high building throws a ball with a speed of ${10\ ms}^{- 1}$ at an angle of $30{^\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $(g = {10\ ms}^{- 2},\sin{30{^\circ}} = 1/2,\ cos\ 30{^\circ} = \sqrt{3}$/2
A) 5.20 m
B) 4.33 m
C) 2.60 m
D) 8.66 m
Explanation
(d) The ball will be at point P when it is at height of 10 m from the ground. So, we have to find distance OP, which can be calculated by directly considering it as a projectile on a level (OX ).
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$OP = R = \frac{u^{2}\sin{2\theta}}{g}$
$= \frac{10^{2}\sin(2 \times 30{^\circ})}{10}$
$= \frac{10\sqrt{3}}{2} = 5\sqrt{3}$
$= 8.66\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.