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,
$T_1=\frac{2u\sin (\alpha -\beta )}{g\cos \beta }$
$=\frac{2u\sin (2\theta -\theta )}{g\cos \theta }$
$=\frac{2u}{g}\tan \theta$
Time of flight of $B$ is, $T_2=\frac{2u\sin \theta }{-g\cos \theta }$
$=\frac{2u}{\tan \theta }$
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 $AB$, therefore collision will take place between the two in mid air.