question_answer

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 \[{{30}^{o}}\] with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? \[[g=10\,m/{{s}^{2}},\sin {{30}^{o}}=1/2,\cos {{30}^{o}}=\sqrt{3}/2]\] [AIEEE 2003]

A) \[5.20\] m

B) \[4.33\] m

C) \[2.60\] m

D) \[8.66\] m

Correct Answer: D

Solution :

[d] The ball will be at point P when it is at a height of 10m from the ground. So, we have to find the distance OP, which can be calculated directly by considering it as a projectile on a levelled plane (OX)-

\[\therefore \,\,\,\,\,OP=R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{10}^{2}}\times \sin (2\times {{30}^{o}})}{10}\]

\[=\frac{10\sqrt{3}}{2}=5\sqrt{3}=8.66\,\,m\]