Question

Airplanes $A$ and $B$ are flying with constant velocity in the same vertical plane at angles $30^{°}$ and $60^{°}$ with respect to the horizontal as shown in figure. The speed of $A$ is $100\sqrt{3}m{s}^{-1}$. At time $t=0s$, an observer in $A$ finds $B$ at a distance of $500m$. This observer sees $B$ moving with a constant velocity perpendicular to the line of motion of $A$. If at $t=t_0$, $A$ just escapes being hit by $B$, $t_0$ is

• A. $3s$
• B. $5s$
• C. $2s$
• D. $1s$

Answer 1

Answer: (B)

The relative velocity of $B$ with respect to $A$ is perpendicular to the line of motion of $A$.
$\therefore v_{B}cos30^{°}=v_A$
$v_B=\frac{v_A}{cos30^{°}}$
$=\frac{100\sqrt{3}}{\sqrt{3}/2}200m{s}^{-1}$
$\therefore t_0=\frac{500}{v_{B}sin30^{\circ }}$
$=\frac{500}{200\times \frac{1}{2}}=5s$