An artillery shell is launched on a flat, horizontal field at an angle of α = 31.5° with respect to the horizontal and with an initial speed of v0 = 266 m/s. What is the horizontal distance covered by the shell after 9.44 s of flight?
To find the horizontal distance covered by the shell, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Initial speed, v0 = 266 m/s
Launch angle, α = 31.5°
Time of flight, t = 9.44 s
First, we can calculate the horizontal component of the initial velocity (v0x) using the formula: v0x = v0 * cos(α).
v0x = 266 m/s * cos(31.5°)
v0x ≈ 230.34 m/s
Next, we can calculate the horizontal distance covered by the shell (x), which can be found using the formula: x = v0x * t.
x = 230.34 m/s * 9.44 s
x ≈ 2175.94 meters
Therefore, the horizontal distance covered by the shell after 9.44 seconds of flight is approximately 2175.94 meters.