JEE MAIN - Physics (2003 - No. 55)
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?
$$\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]$$
5.20 m
4.33 m
2.60 m
8.66 m
Explanation
From the figure it is clear that maximum horizontal range
$$R = {{{u^2}\sin 2\theta } \over g}$$
$$ = {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$$
$$ = 5\sqrt 3 $$ = 8.66 m
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