A student is able to throw a ball vertically to maximum height of 40m. The maximum distance to which the student can throw the ball in the horizonal direction:-

Options

A. `40 (2)^(1//2)m`
B. `20(2)^(1//2)m`
C. 20m
D. 80m

Explanation

To solve the problem, we need to determine the maximum horizontal distance (range) a student can throw a ball, given that the maximum height achieved when thrown vertically is 40 meters.

### Step-by-Step Solution:

1. **Understanding Maximum Height**:
The maximum height (H) achieved by a projectile is given by the formula:
\[ H = \frac{v^2 \sin^2 \theta}{2g} \]
where \(v\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

2. **Setting Up the Equation**:
Since the maximum height is given as 40 meters, we can set up the equation:
\[ 40 = \frac{v^2 \sin^2 90^\circ}{2g} \]
Here, \(\sin 90^\circ = 1\), so the equation simplifies to:
\[ 40 = \frac{v^2}{2g} \]

3. **Solving for Initial Velocity (v)**:
Rearranging the equation to solve for \(v^2\):
\[ v^2 = 80g \]
Substituting \(g = 9.81 \, \text{m/s}^2\):
\[ v^2 = 80 \times 9.81 = 784.8 \, \text{m}^2/\text{s}^2 \]

4. **Calculating the Maximum Range**:
The range (R) of a projectile is given by the formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
The maximum range occurs when \(\theta = 45^\circ\), where \(\sin 2\theta = \sin 90^\circ = 1\). Thus, the formula simplifies to:
\[ R = \frac{v^2}{g} \]

5. **Substituting Values**:
Now substituting \(v^2 = 784.8 \, \text{m}^2/\text{s}^2\) into the range formula:
\[ R = \frac{784.8}{9.81} \approx 80 \, \text{m} \]

6. **Final Answer**:
Therefore, the maximum distance to which the student can throw the ball in the horizontal direction is approximately **80 meters**.