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. 125m
B. 50m
C. 100m
D. 150m

Explanation

To solve the problem, we need to find the maximum height attained by a projectile given the time of flight and the range. Let's break down the steps systematically.

### Step-by-Step Solution:

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

2. **Use the Formula for Time of Flight:**
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Rearranging this formula gives us:
\[ u \sin \theta = \frac{gT}{2} \]
Substituting the known values:
\[ u \sin \theta = \frac{10 \times 10}{2} = 50 \text{ m/s} \quad \text{(Equation 1)} \]

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 \(\sin 2\theta\) as \(2 \sin \theta \cos \theta\):
\[ R = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \]
Rearranging gives:
\[ u^2 \sin 2\theta = \frac{Rg}{2} \]
Substituting the known values:
\[ u^2 \sin 2\theta = \frac{500 \times 10}{2} = 2500 \quad \text{(Equation 2)} \]

4. **Relate \(\sin 2\theta\) to \(\sin \theta\):**
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
From Equation 1, we can express \(u \sin \theta\):
\[ u \sin \theta = 50 \implies u^2 \sin^2 \theta = 2500 \]
Therefore, we can write:
\[ u^2 \sin 2\theta = 2 \cdot 2500 \cdot \cos \theta \]
Substituting into Equation 2 gives:
\[ 2500 \cos \theta = 2500 \implies \cos \theta = 1 \quad \text{(This means } \theta = 0\text{, which is not possible)} \]

5. **Calculate Maximum Height:**
The maximum height (h) attained by the projectile is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
From Equation 1, we know:
\[ u^2 \sin^2 \theta = (u \sin \theta)^2 = 50^2 = 2500 \]
Substituting this into the height formula:
\[ h = \frac{2500}{2 \times 10} = \frac{2500}{20} = 125 \text{ meters} \]

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