Question

A cannon ball is fired with an initial velocity of 85. m/s at an angle of 45° above the horizontal. What is the horizontal distance traveled during the flight?

Answer 1

To calculate the horizontal distance traveled by a cannonball fired at an initial velocity of 85 m/s at an angle of 45° above the horizontal, we can use the following steps:

1. **Resolve the initial velocity into horizontal and vertical components**:
- The horizontal component (\(v_{x0}\)) is given by:
\[
v_{x0} = v_0 \cdot \cos(\theta)
\]
- The vertical component (\(v_{y0}\)) is given by:
\[
v_{y0} = v_0 \cdot \sin(\theta)
\]

where \(v_0 = 85 \, \text{m/s}\) and \(\theta = 45°\).

Calculating these components:
- Since \(\cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2}\):
\[
v_{x0} = 85 \cdot \frac{\sqrt{2}}{2} \approx 60.1 \, \text{m/s}
\]
\[
v_{y0} = 85 \cdot \frac{\sqrt{2}}{2} \approx 60.1 \, \text{m/s}
\]

2. **Determine the time of flight**:
The time of flight (\(t\)) for a projectile can be found using the formula:
\[
t = \frac{2 \cdot v_{y0}}{g}
\]
where \(g \approx 9.81 \, \text{m/s}^2\) is the acceleration due to gravity.

Substituting in the values:
\[
t = \frac{2 \cdot 60.1}{9.81} \approx \frac{120.2}{9.81} \approx 12.24 \, \text{s}
\]

3. **Calculate the horizontal distance traveled**:
The horizontal distance (\(d\)) can be calculated using:
\[
d = v_{x0} \cdot t
\]
Substituting in the values:
\[
d = 60.1 \cdot 12.24 \approx 735.7 \, \text{m}
\]

Therefore, the horizontal distance traveled during the flight is approximately **735.7 meters**.