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

Options

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

Answer

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

### Step 1: Identify the formulas
We know the following formulas for projectile motion:
1. Horizontal Range (R) = \(\frac{u^2 \sin 2\theta}{g}\)
2. Maximum Height (H) = \(\frac{u^2 \sin^2 \theta}{2g}\)

Where:
- \(u\) = initial velocity
- \(\theta\) = angle of projection
- \(g\) = acceleration due to gravity (approximately \(10 \, m/s^2\))

### Step 2: Set up the equations
Given:
- Horizontal Range \(R = 24\)
- Maximum Height \(H = 8\)

Using the formulas:
1. \(\frac{u^2 \sin 2\theta}{g} = 24\) (1)
2. \(\frac{u^2 \sin^2 \theta}{2g} = 8\) (2)

### Step 3: Rearranging the equations
From equation (1):
\[ u^2 \sin 2\theta = 24g \]
From equation (2):
\[ u^2 \sin^2 \theta = 16g \]

### Step 4: Divide the equations
Now, we can divide equation (1) by equation (2):
\[ \frac{u^2 \sin 2\theta}{u^2 \sin^2 \theta} = \frac{24g}{16g} \]
This simplifies to:
\[ \frac{\sin 2\theta}{\sin^2 \theta} = \frac{24}{16} = \frac{3}{2} \]

Using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ \frac{2 \sin \theta \cos \theta}{\sin^2 \theta} = \frac{3}{2} \]
This simplifies to:
\[ \frac{2 \cos \theta}{\sin \theta} = \frac{3}{2} \]
Thus:
\[ \frac{\cos \theta}{\sin \theta} = \frac{3}{4} \]
This means:
\[ \tan \theta = \frac{4}{3} \]

### Step 5: Find the angle of projection
To find the angle \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Find the initial velocity
Now, substitute \(\sin \theta\) and \(\cos \theta\) into one of the original equations to find \(u\). We know:
\[ \sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5} \]
Substituting into equation (2):
\[ u^2 \left(\frac{4}{5}\right)^2 = 16g \]
\[ u^2 \cdot \frac{16}{25} = 16 \cdot 10 \]
\[ u^2 = \frac{16 \cdot 250}{16} = 250 \]
Thus:
\[ u = \sqrt{250} = 5\sqrt{10} \, m/s \]

### Final Results
- Initial Velocity \(u = 5\sqrt{10} \, m/s\)
- Angle of Projection \(\theta = \tan^{-1}\left(\frac{4}{3}\right)\)