A cannon-ball is fired at 300 m/s at an angle of 60º to the horizontal. The ball hits the ground at __ km from the starting point. (Record your answer in the correct significant digits)



Answer :

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.