Question

A car starts from rest and accelerates at $5{\text{m/s}}^2$. At $t=4\text{s}$, 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 $t=6\text{s}$?

$(g=10{\text{m/s}}^2)$

• A. $20\text{m/s},5{\text{m/s}}^2$
• B. $20\sqrt{2}\text{m/s},10{\text{m/s}}^2$
• C. $20\sqrt{2}\text{m/s},0$
• D. $20\text{m/s},0$

Answer 1

The correct option is B $20\sqrt{2}\text{m/s},10{\text{m/s}}^2$

Velocity of the car at $t=4\text{s}$, using kinematic equation,

$v=u+at$

where $u=0;a=5{\text{m/s}}^2;t=4\text{s}$

$v=0+5\times 4=20\text{m/s}$

The ball is dropped at this instant, this means the horizontal velocity of the ball will be

$v_h=20\text{m/s}$

This velocity will remain constant as there is no horizontal acceleration.

Now, vertical velocity at $t=6\text{s}$, i.e.,$\Delta t=2\text{s}$, using kinematic equation,

$v=u+at$

where $v=v_v;u=u_v=0;a=a_v=10{\text{m/s}}^2;t=\Delta t=2\text{s}$

$v_v=u_v+a_v\Delta t=0+10\times 2=20\text{m/s}$

Further, the net velocity is,

$v_{\text{net}}=\sqrt{v_h^2+v_v^2}=\sqrt{{20}^2+{20}^2}=20\sqrt{2}\text{m/s}$

The motion of the ball is under gravity, so,

$a_{\text{net}}=g=10{\text{m/s}}^2$

Hence, option $\left(\text{A}\right)$ is the correct answer.