Question

The minimum velocity of a projectile is 50% of its maximum velocity. If the minimum velocity of the projectile is $10~\text{m s}^{-1}$ then the time of flight of the projectile is (Acceleration due to gravity = $10~\text{m s}^{-2}$

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

Hint:

For projectile motion, remember minimum velocity occurs at top of trajectory; maximum at launch or impact. Relate using \(v_{min} = \frac{v_{max}}{2}\) and kinematics formula for time of flight.

Answer 1

The Correct Option is C

Let \(v_{max} = 2v_{min} = 20~ms^{-1}\), \(v_{min} = 10~ms^{-1}\). For a projectile, \(v_{max} = \sqrt{u^2 + 2gh}\), \(v_{min} = \sqrt{u^2 - 2gh}\). Using kinematics, time of flight \(T = \frac{2u \sin \theta}{g}\) → solves to \(T = 2\sqrt{3}~s\).