A projectile is projected with velocity of 25 m/s at an angle with `theta` the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of `theta` will be: [use `g=10 m//s^2` ]

Options

A. `1/2sin^(-1)((5t^2)/(4R))`
B. `1/2sin^(-1)((4R)/(5t^2))`
C. `tan^(-1)((4t^2)/5R)`
D. `cot^(-1)(R/(20t^2))`

Answer

To solve the problem, we need to find the angle \( \theta \) at which a projectile is launched such that after a time \( t \), its inclination with the horizontal becomes zero.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A projectile is launched with an initial velocity \( u = 25 \, \text{m/s} \) at an angle \( \theta \) with the horizontal.
- After time \( t \), the projectile reaches its highest point where its vertical component of velocity becomes zero, making its inclination with the horizontal zero.

2. **Vertical Motion Analysis**:
- The vertical component of the initial velocity is given by:
\[ u_y = u \sin \theta = 25 \sin \theta \]
- At the highest point, the vertical velocity is zero. The time taken to reach the highest point can be calculated using the formula:
\[ t = \frac{u_y}{g} = \frac{25 \sin \theta}{10} = 2.5 \sin \theta \]

3. **Time of Flight**:
- The total time of flight \( T \) for the projectile is given by:
\[ T = \frac{2u \sin \theta}{g} = \frac{2 \times 25 \sin \theta}{10} = 5 \sin \theta \]
- The time to reach the highest point is half of the total time of flight:
\[ t = \frac{T}{2} = \frac{5 \sin \theta}{2} \]

4. **Equating the Two Expressions for Time**:
- From the two expressions for \( t \):
\[ 2.5 \sin \theta = \frac{5 \sin \theta}{2} \]
- Cross multiplying gives:
\[ 5 \sin \theta = 5 \sin \theta \]
- This equation is always true, meaning we need to derive the range \( R \) to find \( \theta \).

5. **Horizontal Range Calculation**:
- The horizontal component of the initial velocity is:
\[ u_x = u \cos \theta = 25 \cos \theta \]
- The horizontal range \( R \) can be calculated as:
\[ R = u_x \cdot T = (25 \cos \theta) \cdot (5 \sin \theta) = 125 \sin \theta \cos \theta \]
- Using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ R = \frac{125}{2} \sin(2\theta) \]

6. **Finding \( \theta \)**:
- Rearranging the range equation:
\[ \sin(2\theta) = \frac{2R}{125} \]
- To find \( \theta \):
\[ 2\theta = \sin^{-1}\left(\frac{2R}{125}\right) \]
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{2R}{125}\right) \]

7. **Final Expression**:
- Thus, the angle \( \theta \) can be expressed in terms of the range \( R \):
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{2R}{125}\right) \]