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. 20 m
D. 80 m

Answer

To solve the problem step by step, we need to find the maximum horizontal distance (range) a student can throw a ball given that they can throw it vertically to a maximum height of 40 meters.

### Step 1: Understand the relationship between height and initial velocity
The maximum height (h) reached 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,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Set the angle for maximum height
For vertical motion, the angle \( \theta \) is \( 90^\circ \) (which means the ball is thrown straight up). Therefore, \( \sin 90^\circ = 1 \).

Substituting this into the height formula:
\[ h = \frac{v^2}{2g} \]

### Step 3: Substitute the given height into the equation
We know the maximum height \( h = 40 \, \text{m} \):
\[ 40 = \frac{v^2}{2g} \]

### Step 4: Rearrange to find \( v^2 \)
Multiplying both sides by \( 2g \):
\[ v^2 = 80g \]

### Step 5: Calculate the range for horizontal throw
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
For maximum range, the angle \( \theta \) should be \( 45^\circ \) (which means \( \sin 90^\circ = 1 \)):
\[ R = \frac{v^2}{g} \]

### Step 6: Substitute \( v^2 \) into the range formula
From Step 4, we have \( v^2 = 80g \):
\[ R = \frac{80g}{g} = 80 \, \text{m} \]

### Conclusion
The maximum horizontal distance to which the student can throw the ball is **80 meters**.