A car starts from rest and accelerates at 5 m/s2. At t=4 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 s ?(Take g=10 m/s²)

To find the velocity and acceleration of the ball at $t=6\mathrm{s}$, we need to analyze the motion of the car and the ball in two distinct phases: before and after the ball is dropped.

Phase 1: Motion of the car from $t=0$ to $t=4\mathrm{s}$
The car starts from rest, so its initial velocity $u=0$.
It accelerates at $a=5{\mathrm{m}/\mathrm{s}}^2$.
At $t=4\mathrm{s}$, the velocity of the car $v_x$ is given by the first equation of motion:
$v_x=u+at$
$v_x=0+\left(5\right)\left(4\right)=20\mathrm{m}/\mathrm{s}$

At $t=4\mathrm{s}$, a person sitting in the car drops a ball. At the instant of release, the ball acquires the horizontal velocity of the car due to inertia. Therefore:
- Horizontal velocity of the ball at release, $v_x=20\mathrm{m}/\mathrm{s}$.
- Vertical velocity of the ball at release, $u_y=0$.

Phase 2: Motion of the ball after release ($t=4\mathrm{s}$ to $t=6\mathrm{s}$)
Once dropped, the ball is in free fall. The only force acting on it is gravity pointing vertically downwards. Therefore:
- The horizontal acceleration is $a_x=0$.
- The vertical acceleration is $a_y=g=10{\mathrm{m}/\mathrm{s}}^2$.
Consequently, the net acceleration of the ball at any time after being dropped (including at $t=6\mathrm{s}$) is simply the acceleration due to gravity:
$a=10{\mathrm{m}/\mathrm{s}}^2$

The duration of the ball's motion in the air is:
$\Delta t=6\mathrm{s}-4\mathrm{s}=2\mathrm{s}$

During these 2 seconds:
- The horizontal component of velocity remains constant:
$v_x=20\mathrm{m}/\mathrm{s}$
- The vertical component of velocity increases due to gravity:
$v_y=u_y+g\left(\Delta t\right)$
$v_y=0+\left(10\right)\left(2\right)=20\mathrm{m}/\mathrm{s}$

The magnitude of the resultant velocity $v$ of the ball at $t=6\mathrm{s}$ is:
$v=\sqrt{v_x^2+v_y^2}$
$v=\sqrt{{20}^2+{20}^2}=\sqrt{400+400}=\sqrt{800}=20\sqrt{2}\mathrm{m}/\mathrm{s}$

Thus, the velocity of the ball is 20√2 m/s and its acceleration is 10 m/s².