The maximum height attaine by a projectile is increased by ` 10%` by increasing its speed of projecton, without changing the angle of projection. What will the percentage increase in the horizontal range.

Options

A. (a)0.2
B. (b)15
C. (c)10
D. (d)0.05

Explanation

To solve the problem, we need to analyze the relationship between the maximum height and the horizontal range of a projectile. Let's break it down step by step.

### Step 1: Understand the formulas
The maximum height \( H \) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) = initial speed of projection
- \( \theta \) = angle of projection
- \( g \) = acceleration due to gravity

The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 2: Relate the increase in height to the speed
According to the problem, the maximum height is increased by 10%. Thus, we can express this as:
\[ H' = H + 0.1H = 1.1H \]
If the speed of projection is increased to \( u' \), the new maximum height becomes:
\[ H' = \frac{(u')^2 \sin^2 \theta}{2g} \]
Setting the two expressions for height equal gives:
\[ 1.1H = \frac{(u')^2 \sin^2 \theta}{2g} \]

### Step 3: Express the original height in terms of speed
Substituting the original height formula into the equation:
\[ 1.1 \left(\frac{u^2 \sin^2 \theta}{2g}\right) = \frac{(u')^2 \sin^2 \theta}{2g} \]
Cancelling \( \sin^2 \theta \) and \( 2g \) from both sides:
\[ 1.1u^2 = (u')^2 \]
Taking the square root of both sides gives:
\[ u' = \sqrt{1.1} u \]

### Step 4: Calculate the new range
Now we can find the new range \( R' \) using the new speed \( u' \):
\[ R' = \frac{(u')^2 \sin 2\theta}{g} \]
Substituting \( u' = \sqrt{1.1} u \):
\[ R' = \frac{(1.1u) \sin 2\theta}{g} = 1.1 \cdot \frac{u^2 \sin 2\theta}{g} = 1.1R \]

### Step 5: Calculate the percentage increase in range
The percentage increase in range is given by:
\[ \text{Percentage Increase} = \frac{R' - R}{R} \times 100\% \]
Substituting \( R' = 1.1R \):
\[ \text{Percentage Increase} = \frac{1.1R - R}{R} \times 100\% = \frac{0.1R}{R} \times 100\% = 10\% \]

### Final Answer
The percentage increase in the horizontal range is **10%**.

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