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 \[30{}^\circ \] 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

Correct Answer: D

Solution :

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\[(OX)\].                 \[OP=R=\frac{{{u}^{2}}\sin 2\theta }{g}\]                 \[=\frac{{{10}^{2}}\times \sin (2\times {{30}^{o}})}{10}\]                 \[=\frac{10\sqrt{3}}{2}=5\sqrt{3}\]                 \[=8.66\,\,m\]