Two guns (A and B) are pointed at each other, A upwards at an angle of 30° with horizontal and B at the same angle of depression as shown in figure. The guns are 40 m apart. If the gun A fires (shot) at the velocity of 350 m/s and gun B fires (shot) at the velocity of 300 m/s respectively at the same time. The shots meet at M. What will be the time of meeting after firing?

Concept:

To find the time of meeting of the two shots, we consider the relative motion of the projectiles in the horizontal direction.

Given:

Distance between guns, d = 40 m

Velocity of shot from Gun A, v_A = 350 m/s

Velocity of shot from Gun B, v_B = 300 m/s

Angle of projection, \(\theta = 30^\circ\)

Calculation:

Resolving velocities into horizontal components:

\( v_{Ax} = v_A \cos 30^\circ = 350 \times \frac{\sqrt{3}}{2} = 175\sqrt{3} \) m/s

\( v_{Bx} = v_B \cos 30^\circ = 300 \times \frac{\sqrt{3}}{2} = 150\sqrt{3} \) m/s

The relative velocity in the horizontal direction is:

\( v_{\text{relative}, x} = v_{Ax} + v_{Bx} = 175\sqrt{3} + 150\sqrt{3} = 325\sqrt{3} \) m/s

Using the formula,

\( \text{Time} = \frac{\text{Distance}}{\text{Relative Velocity}} \)

\( t = \frac{40}{325\sqrt{3}} \)

Approximating \(\sqrt{3} \approx 1.732\) ,

\( t = \frac{40}{325 \times 1.732} = \frac{40}{562.9} \approx \frac{4}{65}\) seconds