Question

A body is projected with a velocity of $30\text{m/s}$ at an angle of ${60}^{\circ }$ to the horizontal. Find the magnitude of velocity when the angle between the velocity vector and acceleration vector is ${120}^{\circ }$.

• A. $20\sqrt{3}\text{m/s}$
• B. $\frac{20}{\sqrt{3}}\text{m/s}$
• C. $10\sqrt{3}\text{m/s}$
• D. $\frac{10}{\sqrt{3}}\text{m/s}$

Answer 1

The correct option is C $10\sqrt{3}\text{m/s}$

Let the final velocity be $V$ when the angle between velocity vector and acceleration vector is ${120}^{\circ }$.
Only acceleration acting on the body will be acceleration due to gravity in downward direction as shown in figure.

Hence angle made by velocity vector with horizontal, $\varphi ={120}^{\circ }-{90}^{\circ }={30}^{\circ }$
Since there is no acceleration in horizontal direction, therefore horizontal velocity remains same during the projectile.
$\therefore u_x\cos {60}^{\circ }=V\cos \varphi$
$30\cos {60}^{\circ }=V\cos {30}^{\circ }$
$15=V\times \frac{\sqrt{3}}{2}$
$\Rightarrow V=10\sqrt{3}\text{m/s}$