Q. A boy playing on the roof of a 10 m high building throws a ball with a speed of $10ms^{- 1}$ 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? $\lbrack g = 10ms^{- 2},\sin{30{^\circ} = 1/2,\ cos\ 60{^\circ} = \sqrt{3}/2}\rbrack$
A) 5.20 m
B) 4.33 m
C) 2.60 m
D) 8.66 m
Explanation
(d) We know that $R = \frac{u^{2}\sin{2\ \theta}}{g}$
$= \frac{10 \times 10 \times \sin{60{^\circ}}}{10} = 10 \times \frac{\sqrt{3}}{2}$
$= 5 \times 1.732 = 8.66\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.