A projectile is thrown with speed 40 `ms^(-1)` at angle `theta` from horizontal . It is found that projectile is at same height at 1s and 3s. What is the angle of projection ?

Options

A. `tan^(-1)((1)/(sqrt(2)))`
B. `tan^(-1)((1)/(sqrt(3)))`
C. `tan^(-1)(sqrt(3))`
D. `tan^(-1)(sqrt(2))`

Answer

To solve the problem, we need to find the angle of projection \( \theta \) for a projectile thrown with a speed of 40 m/s, which reaches the same height at 1 second and 3 seconds after being thrown.

### Step-by-Step Solution:

1. **Understanding the Problem**:
The projectile is at the same height at \( t = 1 \) s and \( t = 3 \) s. This means that the vertical displacement at these two times is equal.

2. **Using the Projectile Motion Equations**:
The vertical position \( h \) of the projectile at any time \( t \) can be given by the equation:
\[ h = u_y t - \frac{1}{2} g t^2 \]
where \( u_y = u \sin \theta \) is the initial vertical component of the velocity, and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

3. **Setting Up the Equations**:
- At \( t = 1 \) s:
\[ h_1 = u \sin \theta \cdot 1 - \frac{1}{2} g \cdot (1)^2 \]
\[ h_1 = u \sin \theta - \frac{g}{2} \]
- At \( t = 3 \) s:
\[ h_3 = u \sin \theta \cdot 3 - \frac{1}{2} g \cdot (3)^2 \]
\[ h_3 = 3u \sin \theta - \frac{9g}{2} \]

4. **Equating the Heights**:
Since \( h_1 = h_3 \):
\[ u \sin \theta - \frac{g}{2} = 3u \sin \theta - \frac{9g}{2} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 3u \sin \theta - u \sin \theta = \frac{9g}{2} - \frac{g}{2} \]
\[ 2u \sin \theta = 4g \]
\[ u \sin \theta = 2g \]

6. **Substituting the Values**:
Given \( u = 40 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 40 \sin \theta = 2 \cdot 10 \]
\[ 40 \sin \theta = 20 \]
\[ \sin \theta = \frac{20}{40} = \frac{1}{2} \]

7. **Finding the Angle**:
The angle \( \theta \) for which \( \sin \theta = \frac{1}{2} \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle of projection \( \theta \) is \( 30^\circ \).