Question

Two bodies A and B separated by some distance on the ground are projected simultaneously in opposite directions towards each other. Body B is projected with a velocity of $20\,ms^{-1}$ at an angle of $45^\circ$ with the horizontal and body A is projected at an angle of $30^\circ$ with the horizontal. If the two bodies collide at a time of $\sqrt{2}\,s$ from the beginning of their motion, then the initial distance between the bodies A and B is nearly}

• A. 24.12 m
• B. 48.24 m
• C. 27.32 m
• D. 54.64 m

Hint:

For collision of projectiles launched simultaneously, first equate their vertical coordinates and then use horizontal motion.

Answer 1

The Correct Option is B

Concept: At collision, both projectiles must have the same vertical coordinate at the same instant.

Step 1: Find velocity of A using vertical motion.
For B, \[ u_{By}=20\sin45^\circ=10\sqrt2 \] For A, \[ u_{Ay}=u_A\sin30^\circ \] Since collision occurs at same height and same time, \[ u_A\sin30^\circ=10\sqrt2 \] \[ u_A=20\sqrt2 \]

Step 2: Calculate horizontal components.
\[ u_{Ax}=20\sqrt2\cos30^\circ \] \[ =10\sqrt6 \] \[ u_{Bx}=20\cos45^\circ=10\sqrt2 \]

Step 3: Distance covered horizontally till collision.
\[ D=(u_{Ax}+u_{Bx})t \] \[ =(10\sqrt6+10\sqrt2)\sqrt2 \] \[ =20(\sqrt3+1) \] \[ =54.64\,m \] \[ \boxed{54.64\,m} \]