Question

A ball is thrown from the top of a tower with an initial velocity of 10 $m{s}^{-1}$ at an angle of 30${}^{\circ }$ with the horizontal If it hits the ground at a distance of 17.3 m from the base of the tower, the height of the tower is (Take g= 10 $ms-2$)

• A. 5 m
• B. 20 m
• C. 15 m
• D. 10 m

Answer 1

The correct option is D 10 m

Given,

The ball is thrown at an angle, $\theta ={30}^o$.

Initial velocity of the ball, $u=10m/s$

Horizontal range of the ball, $R=17.3m$

We know that, $R=ucos\theta t$,

where $t$ is the time of flight

$⟹t=\frac{17.3}{5\sqrt{3}}=2secs$

using equation of motion we get:-

$-h=ut-\frac{1}{2}g{t}^2$

$=10\times 2-\frac{1}{2}10\times 2^2=-10$

$⟹$ Height of tower, $h=10m$