A body is projected at an angle `theta` so that its range is maximum. If T is the time of flight then the value of maximum range is (acceleration due to gravity= g)

Options

A. `(g^2T)/(2)`
B. `(g T)/(2)`
C. `(g T^2)/(2)`
D. `(g^2 T^2)/(2)`

Answer

To find the maximum range of a body projected at an angle \( \theta \), we can follow these steps:

### Step 1: Understand the Range Formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

### Step 2: Determine the Angle for Maximum Range
The range is maximum when \( \theta = 45^\circ \). At this angle:
\[ \sin 2\theta = \sin 90^\circ = 1 \]
Thus, the maximum range \( R_{\text{max}} \) can be simplified to:
\[ R_{\text{max}} = \frac{u^2}{g} \]

### Step 3: Find the Time of Flight
The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting \( \theta = 45^\circ \):
\[ T = \frac{2u \sin 45^\circ}{g} = \frac{2u \cdot \frac{1}{\sqrt{2}}}{g} = \frac{2u}{g\sqrt{2}} \]
Rearranging this gives:
\[ u = \frac{gT\sqrt{2}}{2} \]

### Step 4: Substitute \( u \) into the Maximum Range Formula
Now substitute \( u \) back into the maximum range formula:
\[ R_{\text{max}} = \frac{u^2}{g} \]
Substituting \( u = \frac{gT\sqrt{2}}{2} \):
\[ R_{\text{max}} = \frac{\left(\frac{gT\sqrt{2}}{2}\right)^2}{g} \]
Calculating this gives:
\[ R_{\text{max}} = \frac{g^2 T^2 \cdot 2}{4g} = \frac{gT^2}{2} \]

### Conclusion
Thus, the value of the maximum range is:
\[ R_{\text{max}} = \frac{gT^2}{2} \]

### Final Answer
The correct option is \( \frac{gT^2}{2} \).

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