A player kicks a foot ball at an angle $30^{\circ}$ with the horizontal with an initial speed $30 \mathrm{~ms}^{-1}$. A second player standing at a distance of $21 \sqrt{3} \mathrm{~m}$ from the first and in the direction of kick. Starts running to catch the ball, at the same instant as kicked by first player. The minimum speed of second player to catch the ball before it hits the ground is (Take, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

• A

  $10 \mathrm{~ms}^{-1}$
• B

  $8 \mathrm{~ms}^{-1}$
• Correct answer
  C

  $8 \sqrt{3} \mathrm{~ms}^{-1}$
• D

  $15 \sqrt{3} \mathrm{~ms}^{-1}$

Solution

The motion of the ball is shown below.
$$ \text { Range, } \begin{aligned} R & =\frac{u^2 \sin 2 \theta}{g} \\ & =\frac{900 \sin (2 \times 30)}{10}=\frac{900 \times \sin 60^{\circ}}{10} \\ R & =\frac{900 \times \sqrt{3}}{10 \times 2}=\frac{900 \sqrt{3}}{20}=45 \sqrt{3} \mathrm{~m} \end{aligned} $$ Since, the IInd player is at $21 \sqrt{3} \mathrm{~m}$ in the direction of kick and he is running to catch the ball, so this can be represented as shown
Let the man runs with speed $v$ towards point $P$ to catch the ball just at point $P$ before it touches the ground in the same time when ball reaches the point $P$. The man will travel distance $=24 \sqrt{3} \mathrm{~m}$ $\Rightarrow$ Distance $=$ speed $\times$ time $\Rightarrow \quad 2 u \sqrt{3}=v \times t$ Where, $t$ is the time of run as well as time of flight since both the ball and IInd man start running at same time hence $t=$ time of flight We know that, Time of flight, $t=\frac{2 u \sin \theta}{g}=\frac{2 \times 30 \times \sin 30^{\circ}}{10}$ $$ t=3 \mathrm{~s} $$ Put $t=35$ in Eq. (i) $$ \begin{aligned} & & 2 u \sqrt{3} & =v \times 3 \\ \Rightarrow & & v & =8 \sqrt{3} \mathrm{~m} / \mathrm{s} \end{aligned} $$ Option (c) is the correct answer.

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