A projectile is fired to have maximum range 500 m. Maximum height attained by the projectitle in this case will be

Options

A. 1000 m
B. 500 m
C. 250 m
D. 125 m

Answer

To find the maximum height attained by a projectile that is fired to achieve a maximum range of 500 m, we can follow these steps:

### Step 1: Understand the relationship between range and initial velocity
The formula for maximum range \( R_{max} \) of a projectile is given by:
\[ R_{max} = \frac{u^2 \sin 2\theta}{g} \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \) for this problem).

### Step 2: Set the angle for maximum range
For maximum range, the angle \( \theta \) should be \( 45^\circ \). Therefore, \( \sin 2\theta = \sin 90^\circ = 1 \).

### Step 3: Substitute known values into the range formula
Given that \( R_{max} = 500 \, \text{m} \), we can substitute into the range formula:
\[ 500 = \frac{u^2 \cdot 1}{10} \]

### Step 4: Solve for initial velocity \( u \)
Rearranging the equation gives:
\[ u^2 = 500 \times 10 = 5000 \]
Taking the square root:
\[ u = \sqrt{5000} \approx 70.71 \, \text{m/s} \]

### Step 5: Calculate the maximum height
The formula for maximum height \( h_{max} \) is given by:
\[ h_{max} = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting \( \theta = 45^\circ \) (where \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)):
\[ h_{max} = \frac{u^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}{2g} \]
This simplifies to:
\[ h_{max} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 6: Substitute \( u^2 \) and \( g \)
Now substituting \( u^2 = 5000 \) and \( g = 10 \):
\[ h_{max} = \frac{5000}{4 \times 10} = \frac{5000}{40} = 125 \, \text{m} \]

### Final Answer
The maximum height attained by the projectile is:
\[ \boxed{125 \, \text{m}} \]