The maximum range of a projectile is 22 m. When it is thrown at an angle of `15^(@)` with the horizontal, its range will be-

Options

A. 22m
B. 6m
C. 15m
D. 11m

Answer

To solve the problem of finding the range of a projectile thrown at an angle of \(15^\circ\) with a maximum range of \(22\) m, we can follow these steps:

### Step 1: Understand the formula for the range of a projectile
The range \(R\) of a projectile launched with an initial velocity \(u\) at an angle \(\theta\) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \(g\) is the acceleration due to gravity.

### Step 2: Determine the maximum range
From the problem, we know that the maximum range \(R_{max}\) is \(22\) m. The maximum range occurs at an angle of \(45^\circ\). Therefore, we can express the maximum range as:
\[ R_{max} = \frac{u^2}{g} \]
Setting this equal to \(22\) m gives us:
\[ \frac{u^2}{g} = 22 \]

### Step 3: Calculate the range at \(15^\circ\)
Now, we need to find the range when the projectile is launched at an angle of \(15^\circ\). We can use the range formula:
\[ R = \frac{u^2 \sin(2 \times 15^\circ)}{g} \]
Calculating \(2 \times 15^\circ\):
\[ 2 \times 15^\circ = 30^\circ \]
Now, we know that \(\sin(30^\circ) = \frac{1}{2}\). Thus, we can substitute this value into the range formula:
\[ R = \frac{u^2 \cdot \frac{1}{2}}{g} \]

### Step 4: Substitute \(u^2/g\) from the maximum range
From Step 2, we have \(u^2/g = 22\). Substituting this into our equation gives:
\[ R = \frac{22 \cdot \frac{1}{2}}{1} = 11 \text{ m} \]

### Conclusion
The range of the projectile when thrown at an angle of \(15^\circ\) is \(11\) m.