Question

The initial speed of a projectile fired from ground is $u$ At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$ The time of flight of the projectile is :

• A. $\frac{2 u }{ g }$
• B. $\frac{ u }{2 g }$
• C. $\frac{ u }{ g }$
• D. $\frac{\sqrt{3} u}{g}$

Hint:

The time of flight of a projectile is only dependent on its initial speed and the acceleration due to gravity, and it is independent of the angle of projection if the speed remains the same.

Answer 1

The Correct Option is C

\(u\  cosθ=\frac{\sqrt{3}​u}{2}\)​
\(⇒cosθ=\frac{\sqrt{3}}{2}\)
\(⇒θ=30^∘\)
\(T=\frac{2usin30^∘}{g}​=\frac{u}{g}\)​
Correct answer is option (b) 

Answer 2

At the highest point of the projectile, the vertical component of the velocity becomes zero, and the horizontal component remains unchanged. Given that the speed at the highest point is \( \frac{\sqrt{5}}{2} u \), we use the relation: \[ u \cos \theta = \frac{\sqrt{5}}{2} u \quad \Rightarrow \quad \cos \theta = \frac{\sqrt{5}}{2} \] This implies: \[ \theta = 30^\circ \] Now, using the formula for the time of flight \( T \), which is given by: \[ T = \frac{2u \sin \theta}{g} = \frac{2u \sin 30^\circ}{g} = \frac{u}{g} \] Thus, the time of flight is \( \frac{u}{g} \).