An artillery shell is fired at an angle of 45.3◦ above the horizontal ground with an initial speed of 1710 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min.
Look at the vertical...that is what time it is in the air.
Hf=Hi+vivert*time- 1/2 g time^2
solve for time
vivert=1710*sin45.3 m/s
An artillery shell is fired at an angle of 67.1◦ above the horizontal ground with an initial speed of 1580 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell, neglecting air resistance.
Answer in units of min.
To find the total time of flight of the artillery shell, we can break down the problem into two components: horizontal motion and vertical motion.
1. Horizontal Motion:
Since there is no acceleration in the horizontal direction, the initial horizontal velocity (v₀x) remains constant throughout the motion. The horizontal velocity can be found using the initial speed (v₀) and the launch angle (θ).
v₀x = v₀ * cos(θ)
v₀x = 1710 m/s * cos(45.3°)
v₀x ≈ 1710 m/s * 0.6989
v₀x ≈ 1196.59 m/s
2. Vertical Motion:
In the vertical direction, the acceleration is due to gravity and acts in the downward direction. The vertical velocity (vᵥ) can be found using the initial speed (v₀) and the launch angle (θ).
vᵥ = v₀ * sin(θ)
vᵥ = 1710 m/s * sin(45.3°)
vᵥ ≈ 1710 m/s * 0.7157
vᵥ ≈ 1224.45 m/s
Next, we can use the equation of motion for vertical motion to find the time of flight (t).
y = v₀y * t + (1/2) * g * t²
0 = v₀ * sin(θ) * t - (1/2) * g * t²
Simplifying the equation, we get:
(1/2) * g * t² = v₀ * sin(θ) * t
(1/2) * 9.8 m/s² * t² = 1710 m/s * sin(45.3°) * t
Now we can solve for t:
(4.9) * t² = 1224.45 * t
t² ≈ 1224.45 * t / 4.9
t² ≈ 250.377 * t
t ≈ 250.377 s
Finally, we convert the time to minutes:
t ≈ 250.377 s * (1 min / 60 s)
t ≈ 4.173 min
Therefore, the total time of flight of the shell, neglecting air resistance, is approximately 4.173 minutes.