In a projectile the statement are true or false
`sin2alpha= (gR)/(u^(2))`
where the letters have the usual meanings.

Answer

To determine whether the statement \( \sin 2\alpha = \frac{gR}{u^2} \) is true or false, we will analyze the projectile motion and derive the range formula step by step.

### Step 1: Understanding the Variables
In the given statement:
- \( R \) is the range of the projectile.
- \( \alpha \) is the angle of projection.
- \( u \) is the initial velocity.
- \( g \) is the acceleration due to gravity.

### Step 2: Deriving the Range Formula
The range \( R \) of a projectile launched at an angle \( \alpha \) with initial velocity \( u \) can be derived using the equations of motion.

1. **Vertical Motion**:
The vertical component of the initial velocity is given by:
\[ u_y = u \sin \alpha \]
The time of flight \( T \) can be derived from the vertical motion. The vertical displacement is zero when the projectile returns to the same vertical level:
\[ 0 = u_y T - \frac{1}{2} g T^2 \]
This simplifies to:
\[ T = \frac{2u \sin \alpha}{g} \]

2. **Horizontal Motion**:
The horizontal component of the initial velocity is:
\[ u_x = u \cos \alpha \]
The range \( R \) is given by the horizontal distance traveled during the time of flight:
\[ R = u_x T = (u \cos \alpha) \left(\frac{2u \sin \alpha}{g}\right) \]
Substituting \( u_x \) and \( T \):
\[ R = \frac{2u^2 \sin \alpha \cos \alpha}{g} \]

3. **Using the Double Angle Identity**:
We can use the trigonometric identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
\[ R = \frac{u^2 \sin 2\alpha}{g} \]

### Step 3: Rearranging the Formula
From the derived formula for range \( R \):
\[ R = \frac{u^2 \sin 2\alpha}{g} \]
We can rearrange this to express \( \sin 2\alpha \):
\[ \sin 2\alpha = \frac{gR}{u^2} \]

### Conclusion
The statement \( \sin 2\alpha = \frac{gR}{u^2} \) is **true**.

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