A man can thrown a stone such that it acquires maximum horizontal range 80 m. The maximum height to which it will rise for the same projectile in metre is

Options

A. 10
B. 20
C. 40
D. 50

Explanation

To find the maximum height \( h \) to which a stone can rise when thrown at an angle that gives it a maximum horizontal range of 80 m, we can follow these steps:

### Step 1: Understand the relationship between range and height
For a projectile, the maximum horizontal range \( R \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

### Step 2: Determine the angle for maximum range
The maximum range occurs when \( \sin(2\theta) = 1 \). This happens when:
\[ 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 3: Substitute the known range into the range formula
Given that the maximum horizontal range \( R \) is 80 m, we can substitute this into the range formula:
\[ 80 = \frac{u^2 \cdot 1}{g} \quad \Rightarrow \quad u^2 = 80g \quad \text{(Equation 1)} \]

### Step 4: Use the height formula for maximum height
The maximum height \( h \) for a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
Substituting \( \theta = 45^\circ \):
\[ h = \frac{u^2 \cdot \sin^2(45^\circ)}{2g} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we have:
\[ h = \frac{u^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 5: Substitute \( u^2 \) from Equation 1 into the height formula
Now, substituting \( u^2 = 80g \) into the height formula:
\[ h = \frac{80g}{4g} = \frac{80}{4} = 20 \text{ m} \]

### Conclusion
Thus, the maximum height to which the stone will rise is:
\[ \boxed{20 \text{ m}} \]

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