Maximum range for a projectile motion is given as R, then height will be

Options

A. R
B. 2 R
C. R/4
D. R/2

Answer

To find the maximum height \( h \) of a projectile when the maximum range \( R \) is given, we can follow these steps:

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

### Step 2: Determine the Angle for Maximum Range
The maximum range occurs at an angle of \( 45^\circ \). Therefore, substituting \( \theta = 45^\circ \):
\[ R = \frac{u^2 \sin(90^\circ)}{g} \]
Since \( \sin(90^\circ) = 1 \), we simplify this to:
\[ R = \frac{u^2}{g} \]

### Step 3: Rearranging the Equation for Initial Velocity
From the equation \( R = \frac{u^2}{g} \), we can express \( u^2 \) in terms of \( R \):
\[ u^2 = Rg \]

### Step 4: Use the Formula for Maximum Height
The formula for the maximum height \( h \) of a projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
Substituting \( \theta = 45^\circ \):
\[ h = \frac{u^2 \sin^2(45^\circ)}{2g} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ h = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 5: Substitute \( u^2 \) in the Height Formula
Now, substituting \( u^2 = Rg \) into the height formula:
\[ h = \frac{Rg}{4g} \]
This simplifies to:
\[ h = \frac{R}{4} \]

### Final Answer
Thus, the maximum height \( h \) that can be achieved by the projectile is:
\[ h = \frac{R}{4} \]

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