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 when thrown with a speed of 20 m/s in a tunnel with a ceiling height of 5 m, we can follow these steps:

### Step 1: Understand the motion components
The ball's motion can be broken down into horizontal and vertical components. The initial speed \( u = 20 \, \text{m/s} \) can be divided into:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Determine the maximum height
The maximum height \( H \) reached by the ball can be calculated using the formula:
\[ H = \frac{u_y^2}{2g} \]
Where \( g = 10 \, \text{m/s}^2 \).

### Step 3: Set the maximum height equal to the ceiling height
Since the ball must not hit the ceiling of the tunnel, we set the maximum height \( H \) less than or equal to the ceiling height (5 m):
\[ \frac{(u \sin \theta)^2}{2g} \leq 5 \]

### Step 4: Substitute the values
Substituting the known values:
\[ \frac{(20 \sin \theta)^2}{2 \times 10} \leq 5 \]
This simplifies to:
\[ \frac{400 \sin^2 \theta}{20} \leq 5 \]
\[ 20 \sin^2 \theta \leq 5 \]
\[ \sin^2 \theta \leq \frac{1}{4} \]
Taking the square root:
\[ \sin \theta \leq \frac{1}{2} \]
Thus, \( \theta \) must be less than or equal to \( 30^\circ \).

### Step 5: Calculate the range
The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
To maximize the range, we need to find the maximum value of \( \sin 2\theta \) for \( \theta \leq 30^\circ \).

### Step 6: Find the maximum angle for range
The maximum value of \( 2\theta \) when \( \theta = 30^\circ \) is:
\[ 2\theta = 60^\circ \]
Thus:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 7: Substitute into the range formula
Substituting back into the range formula:
\[ R = \frac{20^2 \cdot \frac{\sqrt{3}}{2}}{10} \]
\[ R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} \]
\[ R = \frac{200\sqrt{3}}{10} \]
\[ R = 20\sqrt{3} \]

### Step 8: Calculate the numerical value
Using \( \sqrt{3} \approx 1.732 \):
\[ R \approx 20 \cdot 1.732 \]
\[ R \approx 34.64 \, \text{m} \]

### Final Answer
The maximum horizontal distance that the ball can travel without hitting the ceiling of the tunnel is approximately **34.64 m**.

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