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 answer is 20√2 m/s, 10 m/s².

Let's break down the problem step-by-step to find the velocity and acceleration of the ball.

Step 1: Understand the state of the ball right before it is dropped.
The car starts from rest, so its initial velocity is $u=0$ m/s. It accelerates horizontally at $a=5$ m/s².
The ball is dropped at $t=4$ s. We first need to find the horizontal velocity of the car (and thus the ball) at this instant. Using the first equation of motion:

$v=u+at$

$v_x=0+\left(5\right)\left(4\right)=20$ m/s.

Since the ball is inside the car, it shares this horizontal velocity. When dropped, no horizontal forces act on it (ignoring air resistance), so its horizontal velocity remains constant at $v_x=20$ m/s.

Step 2: Calculate the vertical velocity of the ball.
Once the ball is dropped, it is in free fall. Its initial vertical velocity is $u_y=0$ m/s. It accelerates downwards due to gravity ($g=10$ m/s²).
We need to find the ball's vertical velocity at $t=6$ s. The time it has been falling is the difference between the final time and the drop time:
$\Delta t=6-4=2$ s.

Using the first equation of motion for the vertical direction:

$v_y=u_y+g\left(\Delta t\right)$

$v_y=0+\left(10\right)\left(2\right)=20$ m/s.

Step 3: Determine the resultant velocity.
Now we have both the horizontal ($v_x$) and vertical ($v_y$) components of the velocity. We can find the magnitude of the resultant velocity using the Pythagorean theorem:

$v=\sqrt{{v_x}^2+{v_y}^2}$

$v=\sqrt{{20}^2+{20}^2}$

$v=\sqrt{400+400}=\sqrt{800}=20\sqrt{2}$ m/s.

Step 4: Determine the acceleration.
Once the ball leaves the person's hand, it is no longer accelerating horizontally with the car. The only force acting on it is the Earth's gravitational pull. Therefore, its acceleration is simply the acceleration due to gravity, which is directed downwards:
$a=g=10$ m/s².

Thus, the velocity of the ball at $t=6$ s is $20\sqrt{2}$ m/s, and its acceleration is $10$ m/s².