Explanation

Velocity of car at $t = 4 s$ = velocity the ball dropped out of a window $v = u + a t$ $\Rightarrow v = 0 + 5 \times 4 = 20 ms^{-1}$ $\therefore$ Velocity of ball along the horizontal and is $20 ms^{-1}$. After 2 seconds of motion: Vertical velocity of ball $(v_{y}) = u_{y} + a_{y}t$ $v_{y} = 0 + 10 \times 2 = 20 ms^{-1} ( \because a_{y} = g = 10 m/s^{2})$ $\therefore$ Magnitude of velocity of ball (v) = $\sqrt{v_{x}^{2} + v_{y}^{2}} = 20\sqrt{2} m/s$ Since the ball is under free fall. Acceleration of ball (a) at $t = 6 s a = g = 10 m/s^{2}$