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 time required for the cannon ball's entire trip up and down?
Answers
GPT-4o mini
To calculate the total time of flight for a cannonball launched at an angle, we can use the following approach:
1. **Resolve the initial velocity into horizontal and vertical components**:
The initial velocity \( V_0 \) is given as 85 m/s and the launch angle \( \theta \) is 45°.
The vertical component of the initial velocity \( V_{0y} \) is given by:
\[
V_{0y} = V_0 \cdot \sin(\theta) = 85 \, \text{m/s} \cdot \sin(45°)
\]
Since \( \sin(45°) = \frac{\sqrt{2}}{2} \):
\[
V_{0y} = 85 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} \approx 60.1 \, \text{m/s}
\]
2. **Calculate the total time of flight using the vertical motion**:
The time to reach the maximum height can be calculated using the formula:
\[
t_{up} = \frac{V_{0y}}{g}
\]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \):
Substituting the values:
\[
t_{up} = \frac{60.1 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx 6.13 \, \text{s}
\]
The total time of flight \( t_{total} \) is twice the time to reach the maximum height:
\[
t_{total} = 2 \cdot t_{up} = 2 \cdot 6.13 \, \text{s} \approx 12.26 \, \text{s}
\]
Therefore, the total time required for the cannonball's entire trip up and down is approximately **12.3 seconds**.
1. **Resolve the initial velocity into horizontal and vertical components**:
The initial velocity \( V_0 \) is given as 85 m/s and the launch angle \( \theta \) is 45°.
The vertical component of the initial velocity \( V_{0y} \) is given by:
\[
V_{0y} = V_0 \cdot \sin(\theta) = 85 \, \text{m/s} \cdot \sin(45°)
\]
Since \( \sin(45°) = \frac{\sqrt{2}}{2} \):
\[
V_{0y} = 85 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} \approx 60.1 \, \text{m/s}
\]
2. **Calculate the total time of flight using the vertical motion**:
The time to reach the maximum height can be calculated using the formula:
\[
t_{up} = \frac{V_{0y}}{g}
\]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \):
Substituting the values:
\[
t_{up} = \frac{60.1 \, \text{m/s}}{9.81 \, \text{m/s}^2} \approx 6.13 \, \text{s}
\]
The total time of flight \( t_{total} \) is twice the time to reach the maximum height:
\[
t_{total} = 2 \cdot t_{up} = 2 \cdot 6.13 \, \text{s} \approx 12.26 \, \text{s}
\]
Therefore, the total time required for the cannonball's entire trip up and down is approximately **12.3 seconds**.