If a projectile having horizontal range of 24 m acquires a maximum height of 8 m, then its initial velocity and the angle of projection are

Options

A. `5sqrt(g), sin^(-1) (0.6)`
B. `5sqrt(g), sin^(-1) (0.8)`
C. `24sqrt(g), sin^(-1) (0.6)`
D. `8sqrt(g), sin^(-1) (0.8)`

Explanation

To solve the problem of finding the initial velocity and angle of projection of a projectile with a horizontal range of 24 m and a maximum height of 8 m, we can follow these steps:

### Step 1: Understand the Relationships
The horizontal range \( R \) and the maximum height \( H \) of a projectile can be expressed using the following equations:
1. \( R = \frac{U^2 \sin 2\theta}{g} \)
2. \( H = \frac{U^2 \sin^2 \theta}{2g} \)

Where:
- \( U \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Substitute Given Values
We know:
- \( R = 24 \, \text{m} \)
- \( H = 8 \, \text{m} \)

### Step 3: Establish the Relationship Between Range and Height
From the equations, we can relate \( R \) and \( H \):
\[ R = 3H \implies 24 = 3 \times 8 \]
This confirms our values are consistent.

### Step 4: Set Up the Equations
Using the equations for \( R \) and \( H \):
1. \( 24 = \frac{U^2 \sin 2\theta}{g} \)
2. \( 8 = \frac{U^2 \sin^2 \theta}{2g} \)

### Step 5: Eliminate \( g \) and \( U^2 \)
From the second equation, we can express \( U^2 \):
\[ U^2 = 16g \sin^2 \theta \]
Substituting \( U^2 \) into the first equation:
\[ 24 = \frac{16g \sin^2 \theta \cdot \sin 2\theta}{g} \]
This simplifies to:
\[ 24 = 16 \sin^2 \theta \cdot 2 \sin \theta \cos \theta \]
\[ 24 = 32 \sin^3 \theta \cos \theta \]

### Step 6: Simplify the Equation
Dividing both sides by 8:
\[ 3 = 4 \sin^3 \theta \cos \theta \]
Rearranging gives us:
\[ \sin^3 \theta \cos \theta = \frac{3}{4} \]

### Step 7: Use the Identity for \( \tan \theta \)
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \tan \theta = \frac{4}{3} \]

### Step 8: Find \( \sin \theta \) and \( \cos \theta \)
From \( \tan \theta = \frac{4}{3} \):
\[ \sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5} \]

### Step 9: Calculate \( U \)
Now substituting \( \sin \theta \) back into the equation for \( H \):
\[ 8 = \frac{U^2 \left(\frac{4}{5}\right)^2}{2g} \]
This leads to:
\[ 8 = \frac{U^2 \cdot \frac{16}{25}}{2g} \]
\[ U^2 = \frac{8 \cdot 2g \cdot 25}{16} = 25g \]
Taking the square root gives:
\[ U = 5\sqrt{g} \]

### Final Results
Thus, the initial velocity \( U \) is \( 5\sqrt{g} \) and the angle of projection \( \theta \) is \( \sin^{-1}\left(\frac{4}{5}\right) \).