Question

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

• A. 50 s
• B. 20 s
• C. 25 s
• D. 12.5 s

Answer 1

The Correct Option is C

The two boys meet at $C$ after a time $t$.
Taking horizontal motion of boy at $A$ from $A$ to $C, 200 = (10 \cos \theta)t$
or $t =\frac{200}{10 \cos \theta} = \frac{200}{10\left(\frac{\sqrt{\left(10\right)^{2} -\left(6\right)^{2}}}{10}\right)} $
$= \frac{200}{\sqrt{\left(10\right)^{2} - \left(6\right)^{2}}} = \frac{200}{8} = 25 s$