The ceiling of a tunnel is 5 m high. What is the maximum horizontal distance that a ball thrown with a speed of `20 ms^(-1)` can go without hitting the ceiling of the tunnel `(g= 10ms^(-2))`

Options

A. `10sqrt3m`
B. `20 sqrt3m`
C. `30 sqrt3`
D. `40 sqrt3m`

Explanation

To solve the problem of finding the maximum horizontal distance a ball can travel without hitting the ceiling of a tunnel that is 5 m high, we can follow these steps:

### Step 1: Understand the Problem
We need to determine the maximum horizontal distance (range) a ball can travel when thrown with a speed of 20 m/s, ensuring that it does not exceed a height of 5 m.

### Step 2: Identify the Relevant Equations
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \( u \) = initial velocity (20 m/s)
- \( \theta \) = angle of projection
- \( g \) = acceleration due to gravity (10 m/s²)

The maximum height \( H \) reached by the projectile is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]

### Step 3: Set the Maximum Height Condition
Since the tunnel's height is 5 m, we set the maximum height equal to 5 m:
\[ \frac{u^2 \sin^2(\theta)}{2g} \leq 5 \]

### Step 4: Substitute Known Values
Substituting \( u = 20 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ \frac{(20)^2 \sin^2(\theta)}{2 \times 10} = \frac{400 \sin^2(\theta)}{20} = 20 \sin^2(\theta) \leq 5 \]

### Step 5: Solve for \( \sin^2(\theta) \)
Rearranging the inequality:
\[ \sin^2(\theta) \leq \frac{5}{20} = \frac{1}{4} \]
Taking the square root:
\[ \sin(\theta) \leq \frac{1}{2} \]

### Step 6: Determine the Angle \( \theta \)
The maximum value of \( \sin(\theta) = \frac{1}{2} \) corresponds to:
\[ \theta = 30^\circ \]

### Step 7: Calculate the Range
Now we can calculate the range using \( \theta = 30^\circ \):
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Calculating \( \sin(2\theta) \):
\[ \sin(2 \times 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Now substituting back into the range formula:
\[ R = \frac{(20)^2 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{400\sqrt{3}}{20} = 20\sqrt{3} \, \text{m} \]

### Final Answer
The maximum horizontal distance that the ball can go without hitting the ceiling of the tunnel is:
\[ \boxed{20\sqrt{3} \, \text{m}} \]