The time of flight of a projectile is 10 s and range is 500m. Maximum height attained by it is [`g=10 m//s^(2)`]

Options

A. `25 m`
B. `50 m`
C. `82 m`
D. `125 m`

Explanation

To solve the problem, we need to find the maximum height attained by a projectile given the time of flight and the range. Here are the steps to derive the solution:

### Step 1: Understand the given data
- Time of flight (T) = 10 seconds
- Range (R) = 500 meters
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Use the formula for time of flight
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Where:
- \( u \) = initial velocity
- \( \theta \) = angle of projection

Rearranging the formula to find \( u \sin \theta \):
\[ u \sin \theta = \frac{gT}{2} \]
Substituting the known values:
\[ u \sin \theta = \frac{10 \times 10}{2} = 50 \, \text{m/s} \]

### Step 3: Use the formula for range
The range of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We can express \( \sin 2\theta \) in terms of \( \sin \theta \):
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Thus, we can rewrite the range formula as:
\[ R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} \]
Substituting \( u \sin \theta = 50 \):
\[ R = \frac{u \cdot 50 \cdot 2 \cos \theta}{g} \]
Now substituting \( R = 500 \) and \( g = 10 \):
\[ 500 = \frac{u \cdot 50 \cdot 2 \cos \theta}{10} \]
This simplifies to:
\[ 500 = 5u \cos \theta \]
Thus:
\[ u \cos \theta = 100 \]

### Step 4: Find \( u \)
Now we have two equations:
1. \( u \sin \theta = 50 \)
2. \( u \cos \theta = 100 \)

We can find \( u \) by squaring both equations and adding them:
\[ (u \sin \theta)^2 + (u \cos \theta)^2 = 50^2 + 100^2 \]
\[ u^2 (\sin^2 \theta + \cos^2 \theta) = 2500 + 10000 \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ u^2 = 12500 \]
Thus:
\[ u = \sqrt{12500} = 111.8 \, \text{m/s} \]

### Step 5: Find the maximum height
The maximum height \( H \) is given by the formula:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
Substituting \( u \sin \theta = 50 \):
\[ H = \frac{50^2}{2 \times 10} = \frac{2500}{20} = 125 \, \text{m} \]

### Final Answer
The maximum height attained by the projectile is **125 meters**.

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