The ceiling of a hall is 40m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of `56ms^(-1)` without hitting the ceiling of the hall is

Options

A. `25^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Explanation

To solve the problem of finding the angle at which a ball can be thrown to achieve maximum horizontal distance without hitting the ceiling of a hall that is 40 meters high, we can follow these steps:

### Step 1: Understand the maximum height formula
The maximum height \( h \) attained by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
where:
- \( u \) is the initial velocity (56 m/s),
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Set up the equation with known values
We know that the maximum height \( h \) should not exceed 40 meters. Therefore, we can set up the equation:
\[ 40 = \frac{56^2 \sin^2 \theta}{2 \times 9.81} \]

### Step 3: Rearrange the equation to solve for \( \sin^2 \theta \)
Rearranging the equation gives:
\[ \sin^2 \theta = \frac{40 \times 2 \times 9.81}{56^2} \]

### Step 4: Calculate the right-hand side
Now we will calculate the right-hand side:
- First, calculate \( 56^2 \):
\[ 56^2 = 3136 \]
- Then calculate \( 40 \times 2 \times 9.81 \):
\[ 40 \times 2 \times 9.81 = 784.8 \]
- Now substitute these values back into the equation:
\[ \sin^2 \theta = \frac{784.8}{3136} \]

### Step 5: Simplify the fraction
Calculating the fraction gives:
\[ \sin^2 \theta = 0.25 \]

### Step 6: Find \( \sin \theta \)
Taking the square root of both sides:
\[ \sin \theta = \frac{1}{2} \]

### Step 7: Determine the angle \( \theta \)
Now, we find the angle \( \theta \) using the inverse sine function:
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \]
This gives:
\[ \theta = 30^\circ \]

### Conclusion
The angle at which the ball may be thrown to achieve maximum horizontal distance without hitting the ceiling is:
\[ \theta = 30^\circ \]

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