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)

The correct option is 20 √ 2 m/s, 10 m/s².

Let's break down the problem step-by-step to understand the motion of the car and the ball.

Step 1: Analyze the motion of the car up to t = 4 s
The car starts from rest, which means its initial velocity is:
$u=0\text{m/s}$
The car accelerates at a constant rate of:
$a=5\text{m/s}{}^2$
At time $t=4\text{s}$, the velocity of the car ($v_{\text{car}}$) can be calculated using the first equation of motion:
$v_{\text{car}}=u+at$
Substituting the given values:
$v_{\text{car}}=0+\left(5\right)\left(4\right)=20\text{m/s}$

Step 2: Analyze the initial conditions of the ball when it is dropped
At the instant $t=4\text{s}$, a person drops a ball out of the car's window.
When an object is dropped from a moving vehicle, it inherits the instantaneous velocity of the vehicle at that moment. Therefore, the ball will have an initial horizontal velocity ($u_x$) equal to the velocity of the car at $t=4\text{s}$:
$u_x=20\text{m/s}$
Since it is dropped vertically, its initial vertical velocity component is:
$u_y=0\text{m/s}$

Step 3: Analyze the acceleration of the ball after release
Once the ball is released, it is no longer in contact with the car. The only force acting on the ball is gravity (neglecting air resistance).
Thus, for any time after $t=4\text{s}$:
- The horizontal acceleration is $a_x=0$.
- The vertical acceleration is due to gravity: $a_y=g=10\text{m/s}{}^2$ downwards.
Therefore, the net acceleration of the ball at $t=6\text{s}$ is simply the acceleration due to gravity:
$a=10\text{m/s}{}^2$

Step 4: Find the velocity of the ball at t = 6 s
We need to determine the velocity components of the ball after it has been in the air for a duration of:
$\Delta t=6\text{s}-4\text{s}=2\text{s}$
- Horizontal component ($v_x$): Since there is no horizontal acceleration, the horizontal velocity remains constant:
$v_x=u_x=20\text{m/s}$
- Vertical component ($v_y$): Using the first equation of motion for vertical direction:
$v_y=u_y+g\Delta t$
$v_y=0+\left(10\right)\left(2\right)=20\text{m/s}$
- Net velocity ($v$): The magnitude of the velocity is the vector sum of its components:
$v=\sqrt{{v_x}^2+{v_y}^2}$
$v=\sqrt{{20}^2+{20}^2}=\sqrt{400+400}=\sqrt{800}=20\sqrt{2}\text{m/s}$

Conclusion:
At $t=6\text{s}$, the velocity of the ball is 20 √ 2 m/s and its acceleration is 10 m/s².