The speed at the maximum height of a projectile is `sqrt(3)/(2)` times of its initial speed 'u' of projection Its range on the horizontal plane:-

Options

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

Answer

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

### Step 1: Understand the given information
We know that the speed at the maximum height of the projectile is given as \( \frac{\sqrt{3}}{2} u \), where \( u \) is the initial speed of projection.

### Step 2: Analyze the speed at maximum height
At the maximum height of a projectile, the vertical component of the velocity becomes zero. The speed at this point is only due to the horizontal component of the velocity, which remains constant throughout the motion.

Let the angle of projection be \( \theta \). The horizontal and vertical components of the initial velocity \( u \) are:
- Horizontal component: \( u \cos \theta \)
- Vertical component: \( u \sin \theta \)

At maximum height, the speed is given as:
\[ \text{Speed at max height} = u \cos \theta = \frac{\sqrt{3}}{2} u \]

### Step 3: Relate the horizontal component to the angle
From the equation \( u \cos \theta = \frac{\sqrt{3}}{2} u \), we can simplify this to:
\[ \cos \theta = \frac{\sqrt{3}}{2} \]

### Step 4: Determine the angle of projection
The angle \( \theta \) that satisfies \( \cos \theta = \frac{\sqrt{3}}{2} \) is:
\[ \theta = 30^\circ \]

### Step 5: Calculate the range of the projectile
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Substituting \( \theta = 30^\circ \):
\[ R = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin 60^\circ}{g} \]

### Step 6: Substitute the value of \( \sin 60^\circ \)
We know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ R = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} \]

### Final Step: Write the final expression for the range
Thus, the range of the projectile is:
\[ R = \frac{\sqrt{3}}{2} \cdot \frac{u^2}{g} \]

### Conclusion
The range of the projectile is \( \frac{\sqrt{3}}{2} \cdot \frac{u^2}{g} \).

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