Car starts from rest and accelerates with acceleration 2 m/s2. Therefore the velocity of car at t = 5s is 10 m/s. When stone is dropped the velocity of car and hence velocity of stone is 10 m/s in horizontal direction. Since at 5s the stone loses the contact from car, therefore, the stone will move uniformly in horizontal direction and accelerate in vertically downward direction due to gravity.
At t = 5·5s or 0·5 s after the stone is dropped, the X and Y component of velocity are
Motion along horizontal direction:
...(1)
Motion along vertical direction:
...(2)
Eliminating t from (1) and (2),
This the required equation of trajectory. Since the equation of trajectory is the equation of parabola, therefore projectile follows parabolic path.
Velocity of projectile
We know that,
Horizontal range,
We have,
∴
Horizontal range,
i.e. the ball hits the ground at a distance of 20m from the position of player P.
Since the player Q is 12m away, therefore, he has to travel a distance 8m to catch the ball in time equal to the time of flight.
Now the time of flight is,
The velocity of player Q is,