એક ટ્રક સ્થિર સ્થિતિમાંથી શરૂ કરીને $2.0\; m s ^{-2}$ ની પ્રવેગિત ગતિ કરે છે. $t=10$ સેકન્ડે ટ્રકની ઉપર ઊભેલી (જમીનથી $6 \,m$ ઊંચાઈએ) એક વ્યક્તિ પથ્થરને પડવા દે છે. $t = 11$ સેકન્ડે પથ્થરના $(a)$ વેગ અને $(b)$ પ્રવેગ કેટલા હશે ? (હવાનો અવરોધ અવગણો.)

$(a)$ $22.36 \,m / s ,$ 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.