Question

Stone A is projected from top of tower with speed $40\sqrt{2}m/s$ at an angle of ${45}^{\circ }$ with horizontal. At the same time stone B is projected from ground with speed $30m/s$ at an angle of ${30}^{\circ }$ with horizontal as shown.


Find the rate of change of relative velocity of A with respect to B after 1 second from the launch.

• A. $0m/s^2$
• B. $20m/s^2$
• C. $5m/s^2$
• D. $10m/s^2$

Answer 1

The correct option is A $0m/s^2$

The relative velocity of stone A with respect to stone B is given as:
${\overrightarrow{v}}_{AB}={\overrightarrow{v}}_A-{\overrightarrow{v}}_B$
The rate of change of this velocity :
$\frac{d{\overrightarrow{v}}_{AB}}{dt}=\frac{d{\overrightarrow{v}}_A}{dt}-\frac{d{\overrightarrow{v}}_B}{dt}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B$

As the stones are experiencing only one acceleration i.e. acceleration due to gravity. So,
${\overrightarrow{a}}_A={\overrightarrow{a}}_B=\overrightarrow{g}$

$\Rightarrow \overrightarrow{a_{rel}}=\overrightarrow{a_{AB}}=\overrightarrow{g}-\overrightarrow{g}=0$