Question

A particle is projected from a point on the surface of a smooth inclined plane simultaneously another particle q is released on the smooth inclined plane from the same position p and q polite after t is equal to 4 seconds the speed of projection of

Answer 1

Relative velocity of a particle P in the frame of Q changes linearly with time during the flight of P.

Answer 2

According to the given data, let us assume that the inclined plan makes an angle $\theta$ from which projectile Q travels a distance s given by $s\quad =\quad ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$ along with distance traveled along the plane $s\quad =\quad \frac { 1 }{ 2 } gsin\theta$ (16)  which will be $s = 8 g sin \theta.$

Thereby if the projectile P makes an angle a, therefore by the equation of motions, we know that the time of flight $T\quad =\quad u\quad sin\frac { a }{ g } cos\theta$, which is given as T = 4 seconds.

Rearranging the equation, we get – $u sin a = 2 g cos \theta$,  

Distance traveled down the inclined plane will be -

                    $s\quad =\quad ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ \Rightarrow  4 u cos a + 8 g sin \theta = 8 g sin \theta\\\Rightarrow cos a = 0 \\\Rightarrow  a\quad =\quad { 90 }^{\circ}\\\Rightarrow thereby\quad since \quad u sin a = 2 g cos \theta,\\ we \quad get\quad u = 2 g cos \theta.$