A cricket ball hits the ground with a speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$ making an angle $30^{\circ}$ with the horizontal. If the coefficient of restitution is 0.2 , the speed of the reflected ball in $\mathrm{m} \mathrm{s}^{-1}$ is

• A

  $\sqrt{22}$
• B

  $\sqrt{15}$
• Correct answer
  C

  $\sqrt{19}$
• D

  $\sqrt{11}$

Solution

We know that, coefficient of restitution, $e=\left(\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }}\right)$ Now, given collision is as shown
As impact is along normal to surface it does not change the horizontal component of ball. Now if $v_{2 y}$ is final vertical component then $v_{2 y}=e . v_{1 y}=0.2 \times 5 \cos 60^{\circ}$ or $\quad v_{2 y}=0.2 \times 5 \times \frac{1}{2}=0.5 \mathrm{~m} / \mathrm{s}$ So components of final velocity of ball are $v_{2 x}=v_{1 x}=5 \sin 60^{\circ}=(5 \sqrt{3}) / 2$ and $v_{2 y}=e \cdot v_{1 y}=1 / 2$ Hence, final speed of ball, $v_f=\sqrt{\left(v_{2 x}\right)^2+\left(v_{1 x}\right)^2}$ $\begin{aligned} & =\sqrt{\left(\frac{5 \sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2} \\ & =\sqrt{\frac{75}{4}+\frac{1}{4}}=\sqrt{\frac{76}{4}}=\sqrt{19} \mathrm{~m} / \mathrm{s} .\end{aligned}$

Hence, the correct answer is option: 'C'.