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.

Concept:
Impulse is defined as the change in momentum of an object. The impulse-momentum theorem states that:
$\mathrm{Impulse}=\Delta p=m(v_{\mathrm{final}}-v_{\mathrm{initial}})$
where:
• m is the mass of the ball (0.5 kg),
• v_initial is the velocity just before colliding with the ground, and
• v_final is the velocity just after bouncing off the ground.

Step 1: Find the velocity just before hitting the ground (v_initial)
When the ball is dropped from a height of 40 m, it accelerates downwards under gravity. Using the equations of motion:
$v_{\mathrm{initial}}^2=u_0^2+2g{h}_1$
Since the ball is dropped from rest ($u_0=0$):
$v_{\mathrm{initial}}=-\sqrt{2g{h}_1}$
Taking the upward direction as positive and the downward direction as negative:
$v_{\mathrm{initial}}=-\sqrt{2\times 9.8\times 40}$
$v_{\mathrm{initial}}=-\sqrt{784}=-28 \mathrm{m/s}$

Step 2: Find the velocity just after leaving the ground (v_final)
After colliding with the ground, the ball rises upwards to a height of 10 m. At its peak height, its final velocity is zero:
$0=v_{\mathrm{final}}^2-2g{h}_2$
Solving for v_final (which points in the positive upward direction):
$v_{\mathrm{final}}=\sqrt{2g{h}_2}$
$v_{\mathrm{final}}=\sqrt{2\times 9.8\times 10}$
$v_{\mathrm{final}}=\sqrt{196}=14 \mathrm{m/s}$

Step 3: Calculate the impulse
Substitute the velocities into the impulse formula:
$\mathrm{Impulse}=m(v_{\mathrm{final}}-v_{\mathrm{initial}})$
$\mathrm{Impulse}=0.5\times (14-(-28\left)\right)$
$\mathrm{Impulse}=0.5\times (14+28)$
$\mathrm{Impulse}=0.5\times 42=21 \mathrm{N s}$

Therefore, the impulse imparted to the ball during its collision with the ground is 21 NS.