A projectile thrown with an initial velocity of `10 ms^(-1)` at an angle with the horizontal, has a range of 5 m. Taking g=10 `ms^(-2)` and neglecting air resistance, what will be the estimated value of `alpha` ?

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `75^(@)`

Explanation

To find the angle of projection (α) for a projectile thrown with an initial velocity of 10 m/s that has a range of 5 m, we can use the range formula for projectile motion. The formula for the range (R) of a projectile is given by:

\[ R = \frac{u^2 \sin(2\alpha)}{g} \]

Where:
- \( R \) = Range of the projectile
- \( u \) = Initial velocity
- \( g \) = Acceleration due to gravity
- \( \alpha \) = Angle of projection

### Step 1: Substitute the known values into the range formula
Given:
- \( R = 5 \, m \)
- \( u = 10 \, m/s \)
- \( g = 10 \, m/s^2 \)

Substituting these values into the range formula:

\[ 5 = \frac{(10)^2 \sin(2\alpha)}{10} \]

### Step 2: Simplify the equation
We can simplify the equation:

\[ 5 = \frac{100 \sin(2\alpha)}{10} \]

This simplifies to:

\[ 5 = 10 \sin(2\alpha) \]

### Step 3: Solve for \( \sin(2\alpha) \)
Now, we can isolate \( \sin(2\alpha) \):

\[ \sin(2\alpha) = \frac{5}{10} = \frac{1}{2} \]

### Step 4: Determine the angles corresponding to \( \sin(2\alpha) = \frac{1}{2} \)
The angles for which the sine function equals \( \frac{1}{2} \) are:

\[ 2\alpha = 30^\circ \quad \text{or} \quad 2\alpha = 150^\circ \]

### Step 5: Solve for \( \alpha \)
Now, we can find \( \alpha \):

1. From \( 2\alpha = 30^\circ \):
\[ \alpha = \frac{30^\circ}{2} = 15^\circ \]

2. From \( 2\alpha = 150^\circ \):
\[ \alpha = \frac{150^\circ}{2} = 75^\circ \]

### Conclusion
The estimated values of \( \alpha \) are:

\[ \alpha = 15^\circ \quad \text{or} \quad \alpha = 75^\circ \]