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 atr resistance.)

$(a)$ $22.36 \,m / s ,$ at an angle of $26.57^{\circ}$ with the motion of the truck

$(b)$ Initial velocity of the truck, $u=0$

Acceleration, $a=2 \,m / s ^{2}$

Time, $t=10\, s$

As per the first equation of motion, final velocity is given as:

$v=u+a t$

$=0+2 \times 10=20 \,m / s$

The final velocity of the truck and hence, of the stone is $20 \,m / s$

At $t=11\,s ,$ the horizontal component $\left(v_{x}\right)$ of velocity, in the absence of air resistance,

remains unchanged, i.e., $v_{x}=20 \,m / s$

The vertical component $\left(v_{y}\right)$ of velocity of the stone is given by the first equation of

motion as: $v_{y}=u+a_{y} \delta t$

Where, $\delta t=11-10=1 \,s$ and $a_{y}= g =10 \,m / s ^{2}$

$-\because v_{y}=0+10 \times 1=10 \,m / s$

The resultant velocity ( $v$ ) of the stone is given as:

$v =\sqrt{v_{x}^{2}+v_{y}^{2}}$

$=\sqrt{20^{2}+10^{2}}=\sqrt{400+100}$

$=\sqrt{500}=22.36 \,m / s$

Let $\theta$ be the angle made by the resultant velocity with the horizontal component of velocity, $v_{x}$

$\therefore \tan \theta=\left(\frac{v_{y}}{v_{x}}\right)$

$\theta=\tan ^{-1}\left(\frac{10}{20}\right)$

$=\tan ^{-1}(0.5)$

$=26.57^{\circ}$

When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is $10 \,m / s ^{2}$ and it acts vertically downward.