Answer:
Explanation:
To determine how far the cannonball travels horizontally before hitting the ground, we can use the following steps:
1. **Horizontal and Vertical Components of Initial Velocity:**
- The initial velocity \( v_0 \) of the cannonball is 300 m/s at an angle of 60º to the horizontal.
- The horizontal component \( v_{0x} \) of the velocity is \( v_{0x} = v_0 \cos \theta \), where \( \theta = 60^\circ \).
\[
v_{0x} = 300 \cos 60^\circ = 300 \times \frac{1}{2} = 150 \text{ m/s}
\]
- The vertical component \( v_{0y} \) of the velocity is \( v_{0y} = v_0 \sin \theta \).
\[
v_{0y} = 300 \sin 60^\circ = 300 \times \frac{\sqrt{3}}{2} = 150 \sqrt{3} \text{ m/s}
\]
2. **Time of Flight:**
- The time \( t \) of flight can be found using the vertical motion equation \( y = v_{0y} t - \frac{1}{2} g t^2 \), where \( y = 0 \) (since it returns to the same height).
\[
0 = 150 \sqrt{3} \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Solving for \( t \):
\[
t \left( 150 \sqrt{3} - \frac{1}{2} \cdot 9.81 \cdot t \right) = 0
\]
\[
t = 0 \quad \text{or} \quad t = \frac{2 \cdot 150 \sqrt{3}}{9.81} \approx 17.19 \text{ s}
\]
3. **Horizontal Distance:**
- The horizontal distance \( R \) traveled by the cannonball is \( R = v_{0x} \cdot t \).
\[
R = 150 \text{ m/s} \cdot 17.19 \text{ s} \approx 2578.5 \text{ m}
\]
4. **Conversion to Kilometers:**
- Converting the distance to kilometers:
\[
R \approx \frac{2578.5 \text{ m}}{1000} = 2.5785 \text{ km}
\]
Therefore, the cannonball hits the ground at approximately **2.58 km** from the starting point.
