कोई ट्रक विरामावस्था से गति आरंभ करके $2.0\, m s ^{-2}$ के समान त्वरण से गतिशील रहता है। $t=10\, s$ पर, ट्रक के ऊपर खड़ा एक व्यक्ति धरती से $6\, m$ की ऊँचाई से कोई पत्थर बाहर गिराता है। $t=11\, s$ पर, पत्थर का $(a)$ वेग, तथा $(b)$ त्वरण क्या है ? (वायु का प्रतिरोध उपेक्षणीय मानिए ।)

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