Question

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/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 m from the ground? $(g=10\,m/{{s}^{2}},\,\sin {{30}^{o}}=1/2,\,\cos {{30}^{o}}=\sqrt{3}/2)$

• A. 5.20m
• B. 4.33m
• C. 2.60m
• D. 8.66m

Answer 1

Answer: (D)

8.66m

Answer 2

The ball will be at point $P$ when it is at a height of $10\,\,m$ from the ground. So, we have to find distance $OP$ , which can be calculated direct considering it as a projectile on a levelled $(OX)$ . $OP=R=\frac{{{u}^{2}}\sin 2\theta }{g}$ $=\frac{{{10}^{2}}\times \sin (2\times {{30}^{o}})}{10}$ $=\frac{10\sqrt{3}}{2}=5\sqrt{3}$ $=8.66\,\,m$