Question

A ball is projected from ground with a speed of 20 $m s^{- 1}$ at an angle of 45⁰ with horizontal. There is a wall of 25 m height at a distance of 10m from the projection point. The ball will hit the wall at a height of

Moderate

Solution

Since, $u \sin θ \times t = 10 m$
$\Rightarrow 20 \sin 45 ° t = 10$
$\Rightarrow$ $t = \frac{10}{20 \sin 45 °} = \frac{1}{\sqrt{2}} s$
$Now, y = (20 \sin 45 °) t - \frac{1}{2} g t^{2}$
$= 20 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} - \frac{1}{2} \times 10 \times \frac{1}{2} = 7.5 m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME