The time of flight of a projectile is related to its horizontal range by the equation `gT^2 = 2R`. The angle of projection is

Options

A. `30^(@)`
B. `tan^(-1)(2)`
C. `sin^(-1)(1/2)`
D. `45^(@)`

Answer

To solve the problem, we need to find the angle of projection (θ) of a projectile given the relationship between the time of flight (T) and the horizontal range (R) expressed by the equation \( gT^2 = 2R \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The equation given is \( gT^2 = 2R \). Here, \( g \) is the acceleration due to gravity, \( T \) is the time of flight, and \( R \) is the horizontal range of the projectile.

2. **Expression for Time of Flight**:
The time of flight \( T \) for a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]

3. **Expression for Horizontal Range**:
The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

4. **Substituting Time of Flight into the Range Equation**:
Substitute the expression for \( T \) into the equation \( gT^2 = 2R \):
\[ g \left(\frac{2u \sin \theta}{g}\right)^2 = 2 \left(\frac{u^2 \sin 2\theta}{g}\right) \]

5. **Simplifying the Left Side**:
Simplifying the left side:
\[ g \cdot \frac{4u^2 \sin^2 \theta}{g^2} = \frac{4u^2 \sin^2 \theta}{g} \]

6. **Setting Both Sides Equal**:
Now we have:
\[ \frac{4u^2 \sin^2 \theta}{g} = \frac{2u^2 \sin 2\theta}{g} \]

7. **Cancelling Common Terms**:
Cancel \( \frac{u^2}{g} \) from both sides:
\[ 4 \sin^2 \theta = 2 \sin 2\theta \]

8. **Using the Double Angle Identity**:
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substitute this into the equation:
\[ 4 \sin^2 \theta = 2 \cdot 2 \sin \theta \cos \theta \]
This simplifies to:
\[ 4 \sin^2 \theta = 4 \sin \theta \cos \theta \]

9. **Dividing by 4**:
Dividing both sides by 4 gives:
\[ \sin^2 \theta = \sin \theta \cos \theta \]

10. **Rearranging the Equation**:
Rearranging gives:
\[ \sin^2 \theta - \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - \cos \theta) = 0 \]

11. **Finding the Angles**:
This gives us two cases:
- \( \sin \theta = 0 \) (which corresponds to \( \theta = 0^\circ \) or \( 180^\circ \), not valid for projectile motion)
- \( \sin \theta = \cos \theta \) which implies \( \theta = 45^\circ \).

### Conclusion:
Thus, the angle of projection \( \theta \) is \( 45^\circ \).