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?
What is the height of the shell at this moment?
horizontal= 266cos31.5*t
h=266sin31.5*t-4.9t^2
To find the horizontal distance covered by the shell after 9.44 seconds of flight, we can use the horizontal component of the initial velocity and the time of flight.
The horizontal component of the initial velocity, vx₀, can be found by multiplying the initial speed, v₀, by the cosine of the launch angle, α.
vx₀ = v₀ * cos(α)
Substituting the given values:
vx₀ = 266 m/s * cos(31.5°)
Now we can use the equation for horizontal distance:
distance = vx₀ * time
Substituting the values:
distance = (266 m/s * cos(31.5°)) * 9.44 s
Calculate the result to find the horizontal distance covered by the shell.
To find the height of the shell at this moment, we need to calculate the vertical displacement. The vertical component of the initial velocity, vy₀, can be found by multiplying the initial speed, v₀, by the sine of the launch angle, α.
vy₀ = v₀ * sin(α)
Substituting the given values:
vy₀ = 266 m/s * sin(31.5°)
The vertical displacement, Δy, can be calculated using the equation:
Δy = vy₀ * time - (1/2) * g * time^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values:
Δy = (266 m/s * sin(31.5°)) * 9.44 s - (1/2) * 9.8 m/s^2 * (9.44 s)^2
Calculate the result to find the height of the shell at this moment.