A teacher is shot out of a cannon. He leaves the barrel at an angle of 25.0o to the ground with a velocity of 45.0 m/s. How many seconds will the teacher be in flight?
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To determine how many seconds the teacher will be in flight, we can use the kinematic equation for vertical motion.
First, we need to break down the initial velocity into its horizontal and vertical components. Since the teacher leaves the barrel at an angle of 25.0 degrees to the ground, we can use trigonometry to find the vertical component of the velocity.
Vertical component of velocity (V₀y) = V₀ * sin(θ)
V₀ = 45.0 m/s (given)
θ = 25.0 degrees (given)
V₀y = 45.0 m/s * sin(25.0 degrees)
V₀y ≈ 19.06 m/s
Using the kinematic equation for vertical motion, we can find the time it takes for the teacher to reach the highest point of their trajectory (where the vertical velocity becomes zero).
Vf = V₀y + gt
Vf = 0 m/s (at highest point)
g = 9.8 m/s² (acceleration due to gravity)
0 m/s = 19.06 m/s - 9.8 m/s² * t
9.8 m/s² * t = 19.06 m/s
t = 19.06 m/s / 9.8 m/s²
t ≈ 1.946 seconds
Since the total flight time includes both the ascent and descent, we can double the time it took to reach the highest point.
Total flight time ≈ 1.946 seconds * 2
Total flight time ≈ 3.892 seconds
Therefore, the teacher will be in flight for approximately 3.892 seconds.