A ball of mass 0.15kg is dropped from a height 10m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly  $\left(\text{g}=10\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)$

We know that
$\left|\mathrm{Im}\text{pulse}\right|=\text{m}\left|\Delta \right|$ where $\Delta \text{\hspace{0.17em}υ}={\text{υ}}_{\text{2}}-{\text{υ}}_{\text{1}}$
Here,  ${\text{υ}}_{\text{2}}$ is velocity reaches to the same height,  ${\text{υ}}_{\text{1}}$  is velocity just before striking to the ground
For case (1), ball droped from the 10m height and strikes to the ground. Now, the velocity of the ball just before striking to the ground is 
${\text{υ}}_1=-\sqrt{2\text{\hspace{0.17em}gh}}$
$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_1=-\sqrt{2\left(10\right)\left(10\right)}$
$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_1=-10\sqrt{2}\text{m}/\text{s}$
For case (2), ball rebounds to the same height. The velocity with which the ball just reaches to the same height
$v_2=\sqrt{2\mathrm{gh}}\Rightarrow v_2=\sqrt{2\left(10\right)\left(10\right)}$
$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_2=10\sqrt{2}\text{m}/\text{s}$
Now, magnitude of the impulse imparted to the ball,
$\left|\text{Impulse}\right|=\text{m}\left|\Delta \right|$
$=0.15\left|10\sqrt{2}-\left(10\sqrt{2}\right)\right|$
$=0.15\left|2\times 10\sqrt{2}\right|$
$=4.2\text{\hspace{0.17em}kg}-\text{m}/\text{s}$