Question

From an inclined plane two particles are projected with same speed at same angle $ \theta , $ one up and other down the plane as shown in figure, which of the following statements is/are correct?

• A. The time of flight of each particle is the same
• B. The particles will collide the plane with same speed
• C. Both the particles strike the plane perpendicularly
• D. The particles will collide in midair if projected simultaneously and time of flight of each parrick is less than the time of collision

Answer 1

Answer: (A)

Here, $\alpha=2 \theta, \beta=\theta$

Time of flight of $A$ is,
$T_{1}=\frac{2 u \sin (\alpha-\beta)}{g \cos \beta}$
$=\frac{2 u \sin (2 \theta-\theta)}{g \cos \theta}$
$=\frac{2 u}{g} \tan \theta$
Time of flight of $B$ is, $T_{2}=\frac{2 u \sin \theta}{-g \cos \theta}$
$=\frac{2 u}{\tan \theta}$
So $T_{1}=T_{2}$ The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between the two is uniform. The relative velocity of $A$ w.r.t. $B$ is towards $A B$, therefore collision will take place between the two in mid air.