Question

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

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

Answer 1

The correct option is A $5$m/s

Given-

$u=20m/s$ , $\theta ={30}^o$ , $t=0.5s$

Initial velocity-

$\overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j}$

$⟹\overrightarrow{u}=u(\cos {30}^o\hat{i}+\sin {30}^o\hat{j})$

$⟹\overrightarrow{u}=20(\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j})$

$\overrightarrow{u}=10\sqrt{3}\hat{i}+10\hat{j}$

acceleration $a=-g\hat{j}=-10\hat{j}$ since acceleration is acting downward

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$

$⟹\overrightarrow{v}=(10\sqrt{3}\hat{i}+10\hat{j})-(10\hat{j}\times 0.5)$

$\overrightarrow{v}=10\sqrt{3}\hat{i}+5\hat{j}$

Change in velocity=$\Delta \overrightarrow{v}=\overrightarrow{v}-\overrightarrow{u}=-5\hat{j}$

Magnitude of change in velocity $=5m/s$

A short method is-

$\Delta \overrightarrow{v}=acceleration\times t=10\times 0.5=5m/s$

Answer-(A)