What is the relationship between time of flight T and horizontal range R?
(where `theta` is angle of projection with the horizontal)

Options

A. `R=(gT)/(tantheta)`
B. `R=(gT^2)/(2tantheta)`
C. `R=(gT^2)/(tantheta)`
D. `R=(gT)/(2 tan theta)`

Answer

To find the relationship between the time of flight \( T \) and the horizontal range \( R \) for a projectile launched at an angle \( \theta \), we can follow these steps:

### Step 1: Write the formulas for time of flight and horizontal range.
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where:
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity.

The horizontal range \( R \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Express \( R \) in terms of \( T \).
To find the relationship between \( T \) and \( R \), we need to manipulate these equations. First, we can square the time of flight equation:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]

### Step 3: Substitute \( u \) from the range equation.
From the range equation, we can express \( u^2 \) in terms of \( R \):
\[ R = \frac{u^2 \sin 2\theta}{g} \implies u^2 = \frac{Rg}{\sin 2\theta} \]

### Step 4: Substitute \( u^2 \) into the \( T^2 \) equation.
Now, substitute \( u^2 \) into the equation for \( T^2 \):
\[ T^2 = \frac{4 \left(\frac{Rg}{\sin 2\theta}\right) \sin^2 \theta}{g^2} \]

### Step 5: Simplify the equation.
This simplifies to:
\[ T^2 = \frac{4Rg \sin^2 \theta}{g^2 \sin 2\theta} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ T^2 = \frac{4Rg \sin^2 \theta}{g^2 \cdot 2 \sin \theta \cos \theta} = \frac{2R \sin \theta}{g \cos \theta} \]

### Step 6: Rearranging the equation.
Now, we can express \( R \) in terms of \( T^2 \):
\[ R = \frac{g T^2 \cos \theta}{2 \sin \theta} \]

### Conclusion
Thus, the relationship between the time of flight \( T \) and the horizontal range \( R \) is:
\[ R = \frac{g T^2 \cos \theta}{2 \sin \theta} \]