For a projectile, the ratio of maximum height reached to the square of flight time is `(g = 10 ms^(-2))`

Options

A. `5:4`
B. `5:2`
C. `5:1`
D. `10:1`

Explanation

To solve the problem of finding the ratio of maximum height reached (H) to the square of flight time (T²) for a projectile, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formulas**:
- The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
- The total flight time (T) of the projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

2. **Calculate T²**:
- To find T², we square the flight time formula:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]

3. **Set up the ratio H/T²**:
- We need to find the ratio of maximum height to the square of flight time:
\[ \frac{H}{T^2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{4u^2 \sin^2 \theta}{g^2}} \]

4. **Simplify the ratio**:
- Cancel out common terms (u² and sin²θ):
\[ \frac{H}{T^2} = \frac{1}{2g} \cdot \frac{g^2}{4} = \frac{g}{8} \]

5. **Substitute the value of g**:
- Given that \( g = 10 \, \text{m/s}^2 \):
\[ \frac{H}{T^2} = \frac{10}{8} = \frac{5}{4} \]

6. **Conclusion**:
- The ratio of maximum height to the square of flight time is:
\[ H : T^2 = 5 : 4 \]

### Final Answer:
The ratio of maximum height reached to the square of flight time is \( 5 : 4 \).

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