A man can throw a stone 80 meters. The maximum height to which it will rise in meters is

Options

A. 10
B. 20
C. 30
D. 40

Answer

To solve the problem step by step, we will determine the maximum height to which a stone can rise when thrown by a man who can throw it to a maximum range of 80 meters.

### Step 1: Understand the Given Information
The maximum range (R) of the stone is given as 80 meters. We also know that the maximum range is achieved when the angle of projection (θ) is 45 degrees.

### Step 2: Use the Range Formula
The formula for the range of a projectile is given by:
\[ R = \frac{U^2 \sin(2\theta)}{g} \]
Where:
- \( R \) = Range
- \( U \) = Initial velocity
- \( g \) = Acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( \theta \) = Angle of projection

Since the angle of projection is 45 degrees, we can simplify \( \sin(2\theta) \):
\[ \sin(90^\circ) = 1 \]

Thus, the range formula simplifies to:
\[ R = \frac{U^2}{g} \]

### Step 3: Substitute the Known Values
We substitute the known values into the formula:
\[ 80 = \frac{U^2}{10} \]

### Step 4: Solve for Initial Velocity (U)
To find \( U^2 \), we rearrange the equation:
\[ U^2 = 80 \times 10 = 800 \]

### Step 5: Use the Maximum Height Formula
The formula for the maximum height (H) of a projectile is given by:
\[ H = \frac{U^2 \sin^2(\theta)}{2g} \]

Again, since \( \theta = 45^\circ \):
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus, \( \sin^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \)

### Step 6: Substitute Values into the Height Formula
Now we can substitute \( U^2 \) and \( g \) into the height formula:
\[ H = \frac{800 \times \frac{1}{2}}{2 \times 10} \]

### Step 7: Simplify the Equation
Calculating the above:
\[ H = \frac{800 \times 0.5}{20} = \frac{400}{20} = 20 \]

### Conclusion
The maximum height to which the stone will rise is **20 meters**.

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