Question

A boy playing on the roof top of a 10m high building throws a ball with a speed of 10m/s at an angle of ${30}_0$ with the horizontal. How far from the throwing point will the ball be at the height of 10m from the ground.
$\sin {30}^0=\frac{1}{2}\cos {30}^0=\frac{\sqrt{3}}{2}$

• 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

The correct option is A

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

OP=R=$\frac{u^{2}sin2\theta }{g}$

=$\frac{{10}^2\times sin(2\times {30}^0)}{10}$

$\frac{10\sqrt{3}}{2}$=$5\sqrt{3}$=$8.66$