The maximum height attained by a projectile is increased by 10%. Keeping the angle of projection constant, what is the percentage increases in the time of flight?

Options

A. `5%`
B. `10%`
C. `20%`
D. `40%`

Explanation

To solve the problem, we need to establish the relationship between the maximum height attained by a projectile and the time of flight. Let's break it down step by step.

### Step 1: Understand the equations for height and time of flight
The maximum height \( h \) attained 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.

The time of flight \( T \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]

### Step 2: Relate height and time of flight
We can express the time of flight in terms of height. First, let's square the time of flight equation:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]

### Step 3: Substitute for \( u^2 \sin^2 \theta \)
From the height equation, we can express \( u^2 \sin^2 \theta \) in terms of \( h \):
\[ u^2 \sin^2 \theta = 2gh \]
Now, substituting this into the squared time of flight equation gives:
\[ T^2 = \frac{4(2gh)}{g^2} = \frac{8h}{g} \]

### Step 4: Express height in terms of time
From the equation \( T^2 = \frac{8h}{g} \), we can express height \( h \) in terms of time \( T \):
\[ h = \frac{g T^2}{8} \]

### Step 5: Analyze the change in height
According to the problem, the maximum height \( h \) is increased by 10%. Therefore, if the original height is \( h \), the new height \( h' \) is:
\[ h' = h + 0.1h = 1.1h \]

### Step 6: Relate the change in height to the change in time
Using the relationship \( h = \frac{g T^2}{8} \), we can find the new time of flight \( T' \) corresponding to the new height \( h' \):
\[ h' = \frac{g (T')^2}{8} \]
Setting the two expressions for height equal gives:
\[ 1.1h = \frac{g (T')^2}{8} \]

### Step 7: Substitute for \( h \)
Substituting \( h = \frac{g T^2}{8} \) into the equation:
\[ 1.1 \left(\frac{g T^2}{8}\right) = \frac{g (T')^2}{8} \]
Cancelling \( \frac{g}{8} \) from both sides yields:
\[ 1.1 T^2 = (T')^2 \]

### Step 8: Solve for the percentage increase in time of flight
Taking the square root of both sides:
\[ T' = \sqrt{1.1} T \]
The percentage increase in time of flight is given by:
\[ \text{Percentage increase} = \left(\frac{T' - T}{T}\right) \times 100\% = \left(\sqrt{1.1} - 1\right) \times 100\% \]

Calculating \( \sqrt{1.1} \):
\[ \sqrt{1.1} \approx 1.0488 \]
Thus,
\[ \text{Percentage increase} \approx (1.0488 - 1) \times 100\% \approx 4.88\% \]

### Conclusion
The percentage increase in the time of flight is approximately **5%**.

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