Question

A particle is projected with a certain velocity at an angle $\alpha$ above the horizontal from the foot of an inclined plane of inclination 30°. If the particle strikes the plane normally then $\alpha$ is equal to:

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

Answer 1

Answer: (A)

$t_{AB}=$ time of flight of projectile

$=\frac{2usin\left(\alpha -30^{\circ }\right)}{gcos30^{\circ }}$

Now component of velocity along the plane becomes zero at point B.



$\theta =ucos\left(\alpha -30^{\circ }\right)-gsin30^{\circ }\times T$

or $u\cos \left(\alpha -30^{\circ }\right)$

$=gsin30^{\circ }\times \frac{2usin\left(\alpha -30^{\circ }\right)}{gcos30^{\circ }}$

or $tan\left(\alpha -30^{\circ }\right)=\frac{cot30^{\circ }}{2}=\frac{\sqrt{3}}{2}$

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