The initial speed of a projectile fired from ground is `u`.At the highest point during its motion,the speed of projectile is `(sqrt(3))/(2) v` The time of flight of the projectile is :

Options

A. `(2u)/(g)`
B. `(u)/(2g)`
C. `(u)/(g)`
D. `(sqrt(3u)/(g)`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the problem
We are given the initial speed of a projectile as \( u \) and the speed at the highest point as \( \frac{\sqrt{3}}{2} u \). We need to find the time of flight of the projectile.

### Step 2: Analyze the velocity at the highest point
At the highest point of its trajectory, the vertical component of the projectile's velocity becomes zero. The horizontal component remains constant throughout the motion. The speed at the highest point is given as:
\[ v = \frac{\sqrt{3}}{2} u \]
This speed is purely horizontal since the vertical component is zero.

### Step 3: Relate the horizontal component to the initial speed
The horizontal component of the initial velocity can be expressed as:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]
Here, \( \theta \) is the angle of projection.

### Step 4: Simplify the equation
Dividing both sides by \( u \) (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
From trigonometry, we know that:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Thus, we can conclude that:
\[ \theta = 30^\circ \]

### Step 5: Calculate the time of flight
The formula for the time of flight \( T \) of a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting \( \theta = 30^\circ \):
\[ T = \frac{2u \sin 30^\circ}{g} \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have:
\[ T = \frac{2u \cdot \frac{1}{2}}{g} = \frac{u}{g} \]

### Final Answer
The time of flight of the projectile is:
\[ T = \frac{u}{g} \]