Question

A particle is projected from the ground with an initial velocity of $20m/s$ at an angle of ${30}^{\circ }$ with horizontal. The magnitude of change in velocity in a time interval from $t=0$ to $t=0.5s$ is
$(g=10m/s^2)$

• A. $5m/s$
• B. $2.5m/s$
• C. $2m/s$
• D. $4m/s$

Answer 1

The correct option is A $5m/s$

Given, $u=20m/s,\theta ={30}^{\circ },t=0.5s$
$\overrightarrow{u}=u_x\hat{i}+u_y\hat{j}$
$\overrightarrow{u}=20\cos {30}^{\circ }\hat{i}+20\sin {30}^{\circ }\hat{j}\overrightarrow{u}=10\sqrt{3}\hat{i}+10\hat{j}$
Let $v$ be the velocity at $t=0.5s\Rightarrow v=v_x\hat{i}+v_y\hat{j}$
Horizontal component of velocity in a projectile motion always remains constant
$\Rightarrow v_x=u_x=20\cos {30}^{\circ }=10\sqrt{3}\hat{i}$
By using first equation of motion, we can say
$v_y=u_y-gt$
$v_y=20\sin {30}^{\circ }-10\left(0.5\right)=5m/s\therefore v=10\sqrt{3}\hat{i}+5\hat{j}$
Change in velocity $\Delta v=v-u=(10\sqrt{3}\hat{i}+5\hat{j})-(10\sqrt{3}\hat{i}+10\hat{j})\Delta v=-5m{s}^{-1}$

$|\Delta v|=5m{s}^{-1}$