Question

A boy playing on the roof of a 10m 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 $(g=10m/s^2,sin{30}^{\circ }=\frac{1}{2},cos{30}^{\circ }=\frac{\sqrt{3}}{2})$

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

Answer 1

The correct option is A

8.66 m

Simply we have to calculate the range of projectile

$R=\frac{u^{2}sin2\theta }{g}=\frac{\left(10{)}^{2}sin\right(2\times {30}^{\circ })}{10}R=5\sqrt{3}=8.66meter$