In a projectile the statement are true or false
`gT^(2)= 2Rtanalpha`
where the letters have the usual meanings.

Answer

To determine whether the statement \( gT^2 = 2R \tan \alpha \) is true or false, we will analyze the components of projectile motion step by step.

### Step 1: Understand the Variables
- \( g \): Acceleration due to gravity.
- \( T \): Time of flight.
- \( R \): Range of the projectile.
- \( \alpha \): Angle of projection.

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

### Step 3: Square the Time of Flight
Now, we square the time of flight:
\[ T^2 = \left(\frac{2u \sin \alpha}{g}\right)^2 = \frac{4u^2 \sin^2 \alpha}{g^2} \]

### Step 4: Formula for Range
The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\alpha}{g} \]

### Step 5: Substitute Range into the Statement
Now, we need to find \( 2R \):
\[ 2R = 2 \cdot \frac{u^2 \sin 2\alpha}{g} = \frac{2u^2 \sin 2\alpha}{g} \]

### Step 6: Use the Identity for \( \sin 2\alpha \)
Recall that \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \). Therefore:
\[ 2R = \frac{2u^2 (2 \sin \alpha \cos \alpha)}{g} = \frac{4u^2 \sin \alpha \cos \alpha}{g} \]

### Step 7: Substitute \( \tan \alpha \)
Now, we know that \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \). Thus:
\[ 2R \tan \alpha = \frac{4u^2 \sin \alpha \cos \alpha}{g} \cdot \frac{\sin \alpha}{\cos \alpha} = \frac{4u^2 \sin^2 \alpha}{g} \]

### Step 8: Compare LHS and RHS
Now we compare the left-hand side (LHS) and right-hand side (RHS) of the original statement:
- LHS: \( gT^2 = g \cdot \frac{4u^2 \sin^2 \alpha}{g^2} = \frac{4u^2 \sin^2 \alpha}{g} \)
- RHS: \( 2R \tan \alpha = \frac{4u^2 \sin^2 \alpha}{g} \)

Since LHS = RHS, we conclude that the statement \( gT^2 = 2R \tan \alpha \) is **true**.

### Final Answer
The statement \( gT^2 = 2R \tan \alpha \) is **True**.

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