Question

A body is projected horizontally from the top of the tower. When the body strikes the ground the velocity is V and direction of motion makes ${30}^{\circ }$ with the horizontal. The velocity of projection is :

• A. $\frac{\sqrt{3}v}{2}$
• B. $\frac{v}{\sqrt{3}}$
• C. $\frac{v}{2}$
• D. $\frac{2v}{\sqrt{3}}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}v}{2}$

The vertical component of the initial velocity say $u_y=0$

The horizontal component of the initial velocity is $u_x$.

Given: The body strikes the ground with $Vm/s$

As given it strikes at an angle ${30}^{\circ },$

According to question;

$Vcos{30}^{\circ }=u_x$;

$=u_x=V\times \frac{\sqrt{3}}{2}$

$=u_x=\frac{\sqrt{3}}{2}V$

The required velocity of projection is $=u_x=\frac{\sqrt{3}}{2}V$

Option (A) is correct.