Question

A particle is projected with some velocity $u$ making an angle ${60}^0$ with horizontal up an inclined plane. The plane makes an angle ${30}^0$ with horizontal. The particle's range on the inclined plane with it's time of flight $T$, is :

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

Answer 1

Answer: (A)

$\frac{uT}{\sqrt{3}}$

Now we will solve the problem in the perspective of the plane.

For perpendicular motion w.r.t the plane:

Velocity = $usin{30}^0$ = $\frac{u}{2}$

Acceleration = $-gsin{60}^0$ =$\frac{-g\sqrt{3}}{2}$

Using $v=u+at$,

$\frac{T}{2}=\frac{u}{\sqrt{3}g}$

Time of flight $T$ = $\frac{2u}{\sqrt{3}g}$

For parallel motion w.r.t the plane

Velocity = $usin{60}^0$ = $\frac{u\sqrt{3}}{2}$

Acceleration = $-gsin{30}^0$ =$\frac{-g}{2}$

Using $s=ut+\frac{1}{2}a{t}^2$

Range on the inclined plane= $\frac{uT}{\sqrt{3}}$