To calculate the total time of flight for a cannonball launched at an angle, we can use the following approach:
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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 \( \sin(45°) = \frac{\sqrt{2}}{2} \): \[ V_{0y} = 85 , \text{m/s} \cdot \frac{\sqrt{2}}{2} \approx 60.1 , \text{m/s} \]
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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.