Question

The angle $\varphi$ which the velocity vector of stone makes with horizontal just before hitting the ground is given by:

• A. tan $\varphi$ = 2 tan $\theta$
• B. tan $\varphi$ = 2 cot $\theta$
• C. tan $\varphi$= $\sqrt{2}tan\theta$
• D. tan $\varphi$ = $\sqrt{2}cot\theta$

Answer 1

The correct option is C tan $\varphi$= $\sqrt{2}tan\theta$

The time taken to reach maximum height and maximum height are

$t=\frac{u\sin \theta }{g}$ and $H=\frac{u^2{\sin }^2\theta }{2g}$

For remaining half, the time of flight is

$t=\sqrt{\frac{\left(2H\right)}{\left(2g\right)}}=\sqrt{\frac{\left(u^2{\sin }^2\theta \right)}{\left(2g^2\right)}}=\frac{\left(9t\right)}{\left(\sqrt{2}\right)}$ Total time of flightiest $+t=t(1+\frac{1}{\sqrt{2}})$

$T=\frac{u\sin \theta }{g}(1+\frac{1}{\sqrt{2}})$

Also horizontal range is

$=u\cos \theta \times T=\frac{u^2\sin 2\theta }{2g}(1+\frac{1}{\sqrt{2}})$

Let $u_y$ and $v_y$ be initial and final vertical components of velocity.

$u_y^2=2gH$ and $v_y^2=2gH$

$v_y=\sqrt{2}uy$

Angle $\left(\varphi \right)$ final velocity makes with horizontal is

$\tan \varphi =\frac{v_y}{u_x}=\sqrt{2}\frac{u_{y-}}{u_x}=\sqrt{2}\tan \theta$