Question

A boy playing on the roof of a $10m$ high building throws a ball with a speed of $10m/s$ at an angle of ${30}^{\circ }$ with the horizontal. The time taken by the ball to cross the point which is at the height of $10m$ from the ground is

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

The correct option is A $1s$

Since, the height of the building and the point is same, the ball will touch the point at a distance $R$ from the point of projection, where $R$ is the Range of projectile.

$\therefore$ $R=\frac{u^2\sin 2\theta }{g}=\frac{\left(10{)}^2\sin \right(2\times {30}^{\circ })}{10}$
$⟹R=5\sqrt{3}\text{m}$
Hence, time taken by the ball to cross the point is
$t=\frac{R}{u\cos \theta }=\frac{5\sqrt{3}}{10\cos \left({30}^{\circ }\right)}$
$⟹t=1s$

Alternate solution :

Since, the height of the building and the point is same, we can say that the time taken by the ball to cross the point is equal to the time of flight of the projectile motion.

$\Rightarrow t=T=\frac{2u\sin \theta }{g}$
$\Rightarrow t=\frac{2\times 10\sin {30}^{\circ }}{10}$
$\Rightarrow t=1\text{s}$