The maximum height reached by projectile is `4 m`. The horizontal range is `12 m`. The velocity of projection in `m s^-1` is (g is acceleration due to gravity)

Options

A. `5 sqrt(g//2)`
B. `3 sqrt(g//2)`
C. `(1)/(3) sqrt(g//2)`
D. `(1)/(5) sqrt(g//2)`

Explanation

To find the velocity of projection for a projectile that reaches a maximum height of 4 m and has a horizontal range of 12 m, we can follow these steps:

### Step 1: Understand the Relationships
The maximum height (H) and the horizontal range (R) of a projectile are related to the initial velocity (u) and the angle of projection (θ) through the following equations:
1. Maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
2. Range:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Express \(\sin 2\theta\)
Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), we can rewrite the range equation:
\[ R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \]

### Step 3: Relate Height and Range
From the maximum height equation, we can express \(u^2\):
\[ u^2 = \frac{2gH}{\sin^2 \theta} \]
Substituting this into the range equation gives:
\[ R = \frac{(2gH/\sin^2 \theta)(2 \sin \theta \cos \theta)}{g} \]
This simplifies to:
\[ R = \frac{4H \cos \theta}{\sin \theta} \]
Thus, we have:
\[ R = 4H \cot \theta \]

### Step 4: Substitute Known Values
Given \(H = 4 \, \text{m}\) and \(R = 12 \, \text{m}\), we can substitute these values:
\[ 12 = 4 \times 4 \cot \theta \]
This simplifies to:
\[ 12 = 16 \cot \theta \]
Thus, we find:
\[ \cot \theta = \frac{12}{16} = \frac{3}{4} \]

### Step 5: Find \(\tan \theta\)
Since \(\cot \theta = \frac{1}{\tan \theta}\), we have:
\[ \tan \theta = \frac{4}{3} \]

### Step 6: Calculate \(\sin \theta\) and \(\cos \theta\)
Using the relationship:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Let \(\sin \theta = 4k\) and \(\cos \theta = 3k\) for some \(k\). Then:
\[ \sin^2 \theta + \cos^2 \theta = 1 \implies (4k)^2 + (3k)^2 = 1 \implies 16k^2 + 9k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5} \]
Thus:
\[ \sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5} \]

### Step 7: Substitute Back to Find \(u\)
Using the maximum height equation:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \implies 4 = \frac{u^2 \left(\frac{4}{5}\right)^2}{2g} \]
This simplifies to:
\[ 4 = \frac{u^2 \cdot \frac{16}{25}}{2g} \implies 4 \cdot 2g = \frac{16u^2}{25} \implies 8g = \frac{16u^2}{25} \]
Multiplying both sides by 25:
\[ 200g = 16u^2 \implies u^2 = \frac{200g}{16} = \frac{25g}{2} \]
Thus:
\[ u = \sqrt{\frac{25g}{2}} = \frac{5\sqrt{g}}{\sqrt{2}} \]

### Final Step: Express in Standard Form
Thus, the velocity of projection \(u\) is:
\[ u = \frac{5\sqrt{g}}{\sqrt{2}} \, \text{m/s} \]