Question

Two particles are separated by a horizontal distance \({}'x'\) as shown. They are projected at the same time as shown with different initial speeds. The time after which the horizontal distance between them becomes zero is
\(\left\{\cos 30^{\circ }=\frac{\sqrt{3}}{2},\,\cos 60^{\circ }=\frac{1}{2}\right\}\)

• A. \(\frac{x}{u}\)
• B. \(\frac{x}{2u}\)
• C. \(\frac{2u}{x}\)
• D. \(\frac{u}{x}\)

Hint:

Horizontal motion is unaffected by gravity, so the gap closes at the sum of the two horizontal speeds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In projectile motion the horizontal velocity stays constant, and the two particles move towards each other horizontally.

Step 2: Calculation
From the figure, the horizontal components of the two velocities add up to a closing speed of \(2u\). Time to close the gap \(x\) is \(t=\frac{x}{2u}\).

Step 3: Compare the options
The vertical components of the two velocities are equal, so the particles stay at the same height and do meet. Only \(\frac{x}{2u}\) has the right dimension of time and the closing speed.

Final Answer:
The horizontal separation becomes zero after \(\frac{x}{2u}\). \[ \boxed{\text{(B) }\dfrac{x}{2u}} \]