Question

A particle is projected from ground with speed 80 m/s at an angle ${30}^0$ with horizontal from ground. The magnitude of average velocity of particle in time interval t = 2 s to t = 6 s is [Take g = 10 $m/s^2$]:

• A. $40\sqrt{2}m/s$
• B. $40m/s$
• C. $40\sqrt{3}m/s$
• D. $80m/s$

Answer 1

The correct option is C $40\sqrt{3}m/s$

The time of flight$=\frac{2u\sin \theta }{g}=\frac{2\times 80\times \sin 30}{10}=8sec$

So, at 4 seconds the body is at the highest point.

At $t=2s$ and $t=6s$ the body is at the same height.

So vertical displacement between $2s$ and $6s$ is $0$ (body went up and returned to the same vertical level)

So, the vertical average velocity for this period is zero.

Horizontal velocity during this period is constant which is equal to$=u\cos \theta =80\times \cos 30=\underset{–}{40\sqrt{3}m/s}$

This is the average velocity between $t=2s$ and $t=6s$