A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s2) nearly

Correct Option: The correct answer is 4.2 kg m/s.

To find the magnitude of the impulse imparted to the ball, we can follow these steps:

Step 1: Find the velocity of the ball just before it strikes the ground.
The ball is dropped from a height of 10 m. Using the third equation of motion, the velocity $v$ of the ball just before collision is given by:
$v^2=u^2+2gh$
Since the ball is dropped from rest, its initial velocity is zero. Thus:
$v=\sqrt{2gh}$
Substituting the given values, where $h=10\text{m}$ and $g=10{\text{m/s}}^2$:
$v=\sqrt{2\times 10\times 10}=\sqrt{200}\approx 14.14\text{m/s}$

Step 2: Find the velocity of the ball just after it rebounds.
The ball rebounds to the same height of 10 m. The magnitude of the velocity $v^{\prime }$ with which it leaves the ground must be equal to the magnitude of the velocity with which it struck the ground:
$v^{\prime }=\sqrt{2gh}\approx 14.14\text{m/s}$

Step 3: Calculate the change in momentum (Impulse).
Let us define the upward direction as positive and the downward direction as negative.
The initial velocity vector is:
$v_i=-14.14\text{m/s}$
The final velocity vector is:
$v_f=+14.14\text{m/s}$
The impulse $J$ is equal to the change in momentum of the ball:
$J=m(v_f-v_i)$
Substituting the mass of the ball $m=0.15\text{kg}$ and the velocities:
$J=0.15\times (14.14-(-14.14\left)\right)$
$J=0.15\times 28.28\approx 4.24\text{kg m/s}$

The magnitude of the impulse imparted to the ball is approximately 4.2 kg m/s.