What is the range of a projectile thrown with velocity `98 m//s` with angle `30^(@)` from horizontal ?

Options

A. `490sqrt(3)m`
B. `245sqrt(3)m`
C. `980sqrt(3)m`
D. 100m

Answer

To find the range of a projectile thrown with an initial velocity of \( 98 \, \text{m/s} \) at an angle of \( 30^\circ \) from the horizontal, we can use the formula for the range of a projectile:

\[ R = \frac{u^2 \sin 2\theta}{g} \]

Where:
- \( R \) is the range,
- \( u \) is the initial velocity,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

### Step 1: Identify the values
- Initial velocity, \( u = 98 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate \( \sin 2\theta \)
First, we need to calculate \( \sin 2\theta \):
\[ \sin 2\theta = \sin(2 \times 30^\circ) = \sin 60^\circ \]
From trigonometric values, we know:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 3: Plug in the values into the range formula
Now we can substitute the values into the range formula:
\[ R = \frac{(98)^2 \cdot \sin 60^\circ}{9.8} \]
\[ R = \frac{(98)^2 \cdot \frac{\sqrt{3}}{2}}{9.8} \]

### Step 4: Calculate \( (98)^2 \)
Calculating \( (98)^2 \):
\[ (98)^2 = 9604 \]

### Step 5: Substitute back into the equation
Now substituting back:
\[ R = \frac{9604 \cdot \frac{\sqrt{3}}{2}}{9.8} \]
\[ R = \frac{9604 \sqrt{3}}{19.6} \]

### Step 6: Simplify the expression
Now, simplifying the expression:
\[ R = 490 \sqrt{3} \, \text{meters} \]

### Final Answer
Thus, the range of the projectile is:
\[ R = 490 \sqrt{3} \, \text{meters} \]