Question

A particle is projected from suitable height from ground with velocity $\overrightarrow{u}=(8\hat{i}+6\hat{j})$. Take y-axis along vertical and x-axis along horizontal. At what time velocity is perpendicular to initial velocity? $(g=10m/s^2)$

• A. $0.2s$
• B. $\frac{3}{7}s$
• C. $\frac{5}{3}s$
• D. $\frac{1}{9}s$

Answer 1

The correct option is C $\frac{5}{3}s$

$\overrightarrow{u}=8\hat{i}+6\hat{j}$

$v_x=8$

$v_y=6-10$

$\overrightarrow{v}=8\hat{i}+\left(6-10t\right)\hat{j}$

$\because \overrightarrow{u}+\overrightarrow{v}are\perp$

$\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=0$

$\Rightarrow \left(8\hat{i}+6\hat{j}\right)\cdot \left(8\hat{i}+\left(6-10t\right)\hat{j}\right)=0$

$\Rightarrow 64+6\left(6-10t\right)=0$

$\Rightarrow 64+36-60t=0$

$\Rightarrow 100=60t$

$\Rightarrow t=\frac{100}{60}=\frac{5}{3}s$

Hence,

option $\left(C\right)$ is correct answer.