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`

Answer

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

### Step 1: Determine the formula for maximum height (h)
The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Determine the formula for time of flight (T)
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]

### Step 3: Calculate the square of the time of flight (T²)
To find the square of the time of flight, we square the time of flight formula:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]

### Step 4: Find the ratio of maximum height to the square of time of flight
Now, we need to find the ratio \( \frac{h}{T^2} \):
\[ \frac{h}{T^2} = \frac{\frac{u^2 \sin^2 \theta}{2g}}{\frac{4u^2 \sin^2 \theta}{g^2}} \]

### Step 5: Simplify the ratio
Substituting the expressions for \( h \) and \( T^2 \):
\[ \frac{h}{T^2} = \frac{u^2 \sin^2 \theta}{2g} \cdot \frac{g^2}{4u^2 \sin^2 \theta} \]
Cancelling \( u^2 \sin^2 \theta \) from the numerator and denominator:
\[ \frac{h}{T^2} = \frac{g}{8} \]

### Step 6: Substitute the value of g
Given that \( g = 10 \, \text{ms}^{-2} \):
\[ \frac{h}{T^2} = \frac{10}{8} = \frac{5}{4} \]

### Conclusion
Thus, the ratio of maximum height reached to the square of flight time is:
\[ \frac{h}{T^2} = \frac{5}{4} \]

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

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