Two boys are standing at ends $A$ and $B$ of a ground where $A B=200 \mathrm{~m}$. The boy at $B$ starts running in a direction perpendicular to $A B$ with a speed of $6 \mathrm{~ms}^{-1}$. The boy at $A$ starts simultaneously with a velocity of $10 \mathrm{~ms}^{-1}$ and catches the other at time, $t$ where the time, $t$ is

• A

  $50 \mathrm{~s}$
• B

  $20 \mathrm{~s}$
• Correct answer
  C

  $25 \mathrm{~s}$
• D

  $12.5 \mathrm{~s}$

Solution

The given situation is shown below

$$ \begin{aligned} &v_{A}=10 \mathrm{~m} / \mathrm{s} \\ &v_{B}=6 \mathrm{~m} / \mathrm{s} \end{aligned} $$ The two boys meet at point $C$ after a time $t$. Horizontal component of velocity of $v_{A}$. $$ \begin{aligned} v_{A B} &=v_{A} \cos \theta=10 \cos \theta \\ &=10\left(\frac{\sqrt{10^{2}-6^{2}}}{10}\right) \\ &=8 \mathrm{~m} / \mathrm{s} \\ \therefore \quad t &=\frac{A B}{v_{A B}}=\frac{200}{8}=25 \mathrm{~s} \end{aligned} $$

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