$A$ truck starts from rest and accelerates uniformly at $2.0 \; m s^{-2}$. At $t = 10 \; s$,a stone is dropped by a person standing on the top of the truck ($6 \; m$ high from the ground). What are the $(a)$ velocity,and $(b)$ acceleration of the stone at $t = 11 \; s$? (Neglect air resistance.)

(N/A) Given:
Initial velocity of the truck,$u = 0$
Acceleration of the truck,$a = 2.0 \; m s^{-2}$
Time at which the stone is dropped,$t = 10 \; s$
$(a)$ Velocity of the stone at $t = 11 \; s$:
At $t = 10 \; s$,the velocity of the truck is $v = u + at = 0 + 2.0 \times 10 = 20 \; m s^{-1}$.
When the stone is dropped,it possesses this horizontal velocity $v_x = 20 \; m s^{-1}$.
In the vertical direction,the stone starts from rest $(u_y = 0)$ and accelerates due to gravity $(g = 10 \; m s^{-2})$. After $\Delta t = 11 - 10 = 1 \; s$,the vertical velocity is $v_y = u_y + g \Delta t = 0 + 10 \times 1 = 10 \; m s^{-1}$.
The resultant velocity $v$ is $\sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.36 \; m s^{-1}$.
The angle $\theta$ with the horizontal is $\tan \theta = \frac{v_y}{v_x} = \frac{10}{20} = 0.5$,so $\theta = \tan^{-1}(0.5) \approx 26.57^{\circ}$.
$(b)$ Acceleration of the stone at $t = 11 \; s$:
Once the stone is dropped,it is no longer in contact with the truck. The only force acting on it is gravity. Therefore,its acceleration is $g = 10 \; m s^{-2}$ acting vertically downwards.