Question

A boy is running along a straight horizontal road with a constant speed \(5\,\text{m s}^{-1}\). While running he throws a stone with a velocity \(30\,\text{m s}^{-1}\) at an angle \(60^\circ\) with the horizontal. Then the time of flight of the stone is: (Given \(g = 10\,\text{m s}^{-2}\))

• A. \(3\sqrt{2}\)
• B. \(\frac{\sqrt{3}}{2}\)
• C. \(4\sqrt{3}\)
• D. \(3\sqrt{3}\)

Hint:

Time of flight depends only on vertical component of velocity, not on horizontal motion or observer’s speed.

Answer 1

The Correct Option is D

Step 1: Understand relative motion.
The boy is moving horizontally, so his velocity only affects the horizontal component of the stone's velocity. The vertical motion remains unaffected.

Step 2: Find vertical component of velocity.
\[ u_y = 30 \sin 60^\circ \]
\[ u_y = 30 \times \frac{\sqrt{3}}{2} \]
\[ u_y = 15\sqrt{3}\,\text{m/s} \]

Step 3: Use formula for time of flight.
\[ T = \frac{2u_y}{g} \]

Step 4: Substitute values.
\[ T = \frac{2 \times 15\sqrt{3}}{10} \]
\[ T = \frac{30\sqrt{3}}{10} \]
\[ T = 3\sqrt{3}\,\text{s} \]

Step 5: Note about horizontal velocity.
The horizontal velocity (including boy’s motion) does not affect time of flight.

Step 6: Interpretation.
Time of flight depends only on vertical motion under gravity.

Step 7: Final answer.
\[ \boxed{3\sqrt{3}} \]