The maximum horizontal range of a projectile is 400 m . The maximum value of height attained by it will be

Options

A. `100 m`
B. `200 m`
C. `400 m`
D. `800 m`

Answer

To find the maximum height attained by a projectile when the maximum horizontal range is given, we can follow these steps:

### Step 1: Understand the relationship between range and height
The horizontal range (R) of a projectile is given by the formula:
\[ R = \frac{U^2 \sin 2\theta}{g} \]
where:
- \( U \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Identify the maximum range condition
The maximum range occurs when \( \sin 2\theta = 1 \), which implies:
\[ 2\theta = 90^\circ \]
Thus:
\[ \theta = 45^\circ \]

### Step 3: Substitute the known range into the range formula
Given that the maximum horizontal range \( R = 400 \, m \), we can substitute this value into the range formula:
\[ 400 = \frac{U^2 \cdot 1}{g} \]
This simplifies to:
\[ U^2 = 400g \]
(Equation 1)

### Step 4: Calculate the maximum height
The maximum height (H) attained by a projectile is given by the formula:
\[ H = \frac{U^2 \sin^2 \theta}{2g} \]
Substituting \( \theta = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus:
\[ H = \frac{U^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{U^2 \cdot \frac{1}{2}}{2g} = \frac{U^2}{4g} \]

### Step 5: Substitute \( U^2 \) from Equation 1 into the height formula
From Equation 1, we have \( U^2 = 400g \). Substituting this into the height formula gives:
\[ H = \frac{400g}{4g} = \frac{400}{4} = 100 \, m \]

### Conclusion
The maximum height attained by the projectile is:
\[ H = 100 \, m \]

### Final Answer
The maximum value of height attained by the projectile is **100 m**.

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