Previous Year NEET Question- Optics L-3

Question: A car starts from rest and accelerates at $5 \mathrm{~m} / \mathrm{s}^2$. At $\mathrm{t}=4 \mathrm{~s}$, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t=6 \mathrm{~s}$ ? $\left(\right.$ Take $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$#

A) $20 \sqrt{2} \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}^2$

B) $20 \mathrm{~m} / \mathrm{s}, 5 \mathrm{~m} / \mathrm{s}^2$

C) $20 \mathrm{~m} / \mathrm{s}, 0$

D) $20 \sqrt{2} \mathrm{~m} / \mathrm{s}, 0$

Answer: $20 \sqrt{2} \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}^2$#

Sol:#

$$ \begin{aligned} & u=0 \ & a=5 \ & t=4 \end{aligned} $$ velocity of car at $t=4 \mathrm{sec}$ is $$ \begin{aligned} & V=u+\text { at } \ & V=0+5 \times 4 \ & V=20 \mathrm{~m} / \mathrm{s} \end{aligned} $$

For ball :

At $t=4 \mathrm{~s}, \mathrm{~A}$ ball is dropped out of a window so velocity of ball at this instant is $20 \mathrm{~ms}-1$ along horizontal.

After 2 seconds of motion :

Horizontal velocity of ball , $\mathrm{V}_{\mathrm{x}}=20 \mathrm{~m} / \mathrm{sec}$ $$ \begin{aligned} & V_y=u+u t \ & =10 \times 2 \end{aligned} $$

Vertical velocity of ball, $V_y=20 \mathrm{~m} / \mathrm{sec}$

So magnitude of velocity of ball $$ \mathrm{V}=\sqrt{V_x^2+V_y^2}=20 \sqrt{2} $$ and Acceleration of ball at $t=6 \mathrm{~s}$ is $\mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^2$ As ball is under free fall.