First slide

Projectile motion

Question

A body is projected horizontally from the top of a tower with initial velocity 18 m s^-1. It hits the ground at angle 45^o. what is the vertical component of velocity when strikes the ground?

Moderate

A

$9 {ms}^{- 1}$
B

$9 \sqrt{2} {ms}^{- 1}$
C

$18 {ms}^{- 1}$
D

$18 \sqrt{2} {ms}^{- 1}$

Solution

$\begin{array}{l} v \cos 45^{\circ} = u = 18 {ms}^{- 1} \\ \Rightarrow v = 18 \sqrt{2} {ms}^{- 1} \end{array}$
Vertical component
$v \sin 45^{\circ} = 18 \sqrt{2} \times \frac{1}{\sqrt{2}} = 18 {ms}^{- 1}$

Get Instant Solutions

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

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App