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 find the maximum height attained by a projectile given the time of flight and range, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Time of flight (T) = 10 seconds
- Range (R) = 500 meters
- Acceleration due to gravity (g) = 10 m/s²

2. **Use the Time of Flight Formula:**
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( u \) is the initial velocity and \( \theta \) is the angle of projection.

3. **Rearranging the Formula:**
We can rearrange the formula to find \( u \sin \theta \):
\[ u \sin \theta = \frac{T \cdot g}{2} \]
Substituting the known values:
\[ u \sin \theta = \frac{10 \cdot 10}{2} = 50 \text{ m/s} \]

4. **Use the Range Formula:**
The range of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We can express \( \sin 2\theta \) as \( 2 \sin \theta \cos \theta \):
\[ R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} \]
Rearranging gives us:
\[ u^2 \sin \theta \cos \theta = \frac{R \cdot g}{2} \]
Substituting the known values:
\[ u^2 \sin \theta \cos \theta = \frac{500 \cdot 10}{2} = 2500 \]

5. **Expressing \( u \cos \theta \):**
From the earlier step, we know:
\[ u \sin \theta = 50 \]
Let \( u \cos \theta = x \). Then we can express \( u^2 \sin \theta \cos \theta \) as:
\[ u^2 \sin \theta \cos \theta = (u \sin \theta)(u \cos \theta) = 50x \]
Setting this equal to 2500 gives:
\[ 50x = 2500 \implies x = 50 \text{ m/s} \]

6. **Finding Maximum Height:**
The maximum height \( h \) is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
We can find \( u^2 \) using:
\[ u^2 = (u \sin \theta)^2 + (u \cos \theta)^2 = 50^2 + 50^2 = 2500 + 2500 = 5000 \]
Now substituting \( u^2 \sin^2 \theta \):
\[ h = \frac{5000 \cdot \left(\frac{50}{u}\right)^2}{2 \cdot 10} \]
Since \( u \sin \theta = 50 \), we can find \( h \):
\[ h = \frac{50^2}{2 \cdot 10} = \frac{2500}{20} = 125 \text{ meters} \]

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

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