question_answer

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

Correct Answer: A

Solution :

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,\[{{T}_{2}}=\frac{2u\sin \theta }{g\cos \theta }\]                                                 \[=\frac{2u}{g}\tan \theta \] 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 midair.