Question

A particle is projected with certain velocity at an angle $\alpha$ above the horizontal from the foot of an inclied plane of inclination ${30}^0$ . If the particle strikes the plane normally then $\alpha$ is

• A. ${30}^0+ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. ${45}^0$
• C. ${60}^0$
• D. ${30}^0+ta{n}^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is A ${30}^0+ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

If the projectile hits the incline normally, it means that component of velocity along the incline is zero.
$v_x=u_x+a_{x}T$

$⟹0=ucos(\alpha -{30}^0)gsin{30}^{0}T$

$T=\frac{ucos(\alpha -{30}^0)}{gsin{30}^0}$

Now, time of flight is also given by the formula :

$⟹T=\frac{2usin(\alpha -{30}^0)}{gcos{30}^0}$

Equating two expressions for time of flight, we have :

$⟹\frac{2usin(\alpha -{30}^0)}{gcos{30}^0}=\frac{ucos(\alpha -{30}^0)}{gsin{30}^0}$

$⟹tan(\alpha -{30}^0)=\frac{1}{2tan{30}^0}=\frac{\sqrt{3}}{2}$

$⟹\alpha ={30}^0+ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$