Initial velocity of car = $0$. Acceleration of car = $5\text{ m/s}^2$. Velocity of car at $t = 4\text{ s}$; $v = u + at \Rightarrow v = 0 + 5 \times 4 = 20\text{ ms}^{-1}$. At $t = 4\text{ s}$, a ball is dropped out of a window so velocity of ball at this instant is $20\text{ ms}^{-1}$ along horizontal. After 2 seconds of motion: Horizontal velocity of ball = $20\text{ ms}^{-1}$ ($\because a_x = 0$). Vertical velocity of ball ($v_y$) = $u_y + a_yt = 0 + 10 \times 2 = 20\text{ ms}^{-1}$ ($\because a_y = g = 10\text{ m/s}^2$). So magnitude of velocity of ball $v = \sqrt{v_x^2 + v_y^2} = 20\sqrt{2}\text{ m/s}$. Acceleration of ball at $t = 6\text{ s}$ is $g = 10\text{ m/s}^2$ as ball is under free fall.