Question

A particle is projected with initial velocity $\overrightarrow{v}=(10\hat{i}+15\hat{j})m/s$ in $x-y$ plane. The magnitude of displacement of the particle at time $t=1s$ is
($\overrightarrow{g}=-10\hat{j}m/s^2$)

• A. $10m$
• B. $10\sqrt{2}m$
• C. $20m$
• D. $20\sqrt{2}m$

Answer 1

Answer: (B)

$10\sqrt{2}m$

Given,

$\overrightarrow{v}=(10\hat{i}+15\hat{j})m/s$

$g=-10\hat{j}m/s^2$

$t=1s$

From 2nd equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$\overrightarrow{s}=(10\hat{i}+15\hat{j})\times 1+\frac{1}{2}(-10\hat{j})(1{)}^2$

$\overrightarrow{s}=10\hat{i}+15\hat{j}-5\hat{j}$

$\overrightarrow{s}=10\hat{i}+10\hat{j}$

Magnitude of the displacement,

$|\overrightarrow{s}|=\sqrt{(10{)}^2+(10{)}^2}=\sqrt{200}m$

$|\overrightarrow{s}|=10\sqrt{2}m$

The correct option is B.