First slide

Projectile motion

Question

A player kicks a ball at a speed of 20 ms^-1 so that its horizontal range is maximum. Another player 24 m away in the direction of kick starts running in the same direction at the same instant of hit. If he has to catch the ball just before it reaches the ground, he should run with a velocity equal to (Take g =10 ms^-2)

Moderate

A

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

$4 \sqrt{2} {ms}^{- 1}$
C

$6 \sqrt{2} m s^{- 1}$
D

$10 \sqrt{2} {ms}^{- 1}$

Solution

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
Horizontal range is maximum when angle of projection $θ$ is 45^o
$R_{max} = \frac{u^{2}}{g} = \frac{20^{2}}{10} = 40 m$
Time of flight, $T = \frac{2 usin 45^{\circ}}{g} = \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} = 2 \sqrt{2} s$
The player can catch the ball before reaching the ground if he covers a distance $= 40 - 24 = 16 m in 2 \sqrt{2} s$
The velocity of that player must be $= \frac{16}{2 \sqrt{2}} = 4 \sqrt{2} {ms}^{- 1}$

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