Question

A ball is thrown with a velocity of $8m{s}^{-1}$ making an angle of ${60}^o$ with the horizontal. It's velocity will be perpendicular to the direction of initial velocity of projection after a time of ($g=10m{s}^{-2}$)

• A. $\frac{1.6}{\sqrt{3}}s$
• B. $\frac{4}{\sqrt{3}}s$
• C. $4\sqrt{3}s$
• D. $0.8\sqrt{3}s$

Answer 1

The correct option is A $\frac{1.6}{\sqrt{3}}s$

$u\left(t\right)=v\cos {60}^{\circ }\hat{i}+(v\sin {60}^{\circ }-gt)\hat{j}$
$u\left(o\right)=v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}$

At time $t$:

$u\left(t\right)\cdot u\left(0\right)=0$ for these to be $\perp$
$\Rightarrow (v\cos {60}^{\circ }{)}^2+(v\sin {60}^{\circ }{)}^2-v\sin {60}^{\circ }gt=0$

$\Rightarrow v^2=v\sin {60}^{\circ }gt$

$\Rightarrow t=\frac{v}{g}\frac{2}{\sqrt{3}}=\frac{8\times 2}{g\sqrt{3}}=\frac{16}{10\times 3}\sqrt{3}$$=\frac{1.6}{\sqrt{3}}$s