A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)

The correct option is 21 Ns.

To find the impulse imparted to the ball during its collision with the ground, we use the impulse-momentum theorem, which states that the impulse ($J$) is equal to the change in momentum of the ball:
$J=\Delta p=p_{\text{final}}-p_{\text{initial}}$

Step 1: Find the velocity of the ball just before it hits the ground ($v_1$)
The ball is dropped from a height $h_1=40\text{m}$. Using the equation of motion $v^2=u^2+2gh$, where the initial velocity $u=0$:
$v_1=\sqrt{2g{h}_1}$
Substituting the given values ($g=9.8{\text{m/s}}^2$ and $h_1=40\text{m}$):
$v_1=\sqrt{2\times 9.8\times 40}=\sqrt{784}=28\text{m/s}$
Taking the downward direction as negative, the velocity just before collision is:
${\overrightarrow{v}}_1=-28\hat{j}\text{m/s}$

Step 2: Find the velocity of the ball just after it rebounds from the ground ($v_2$)
The ball rises to a height $h_2=10\text{m}$. Using the equation of motion for upward motion to maximum height:
$v_2=\sqrt{2g{h}_2}$
Substituting the values ($g=9.8{\text{m/s}}^2$ and $h_2=10\text{m}$):
$v_2=\sqrt{2\times 9.8\times 10}=\sqrt{196}=14\text{m/s}$
Since the rebound direction is upward (positive direction):
${\overrightarrow{v}}_2=+14\hat{j}\text{m/s}$

Step 3: Calculate the impulse imparted to the ball
The mass of the ball is $m=0.5\text{kg}$.
The impulse is:
$J=m(v_2-v_1)$
$J=0.5\times [14-(-28\left)\right]$
$J=0.5\times (14+28)$
$J=0.5\times 42=21\text{Ns}$

Thus, the impulse imparted to the ball during its collision with the ground is 21 Ns.