and s_0 is the initial position. These work for when the acceleration is zero, as in the horizontal component of velocitys(t)_horz = (v_0_horz)t + (s_0_horz)ands(t)_vert = (1/2)at^2 + (v_0_vert)t + (s_0_vert)-Since projectile motion is divided in X and Y components, does that mean I can swap x for y in equations when Im solving for the Y part.........
The equations of motion with constant acceleration, derivable with basic calculus, are
s(t) = (1/2)at^2 + (v_0)t + s_0
v(t) = at + v_0
where s(t) is position as a function of time, v(t) is velocity as a function of time, v_0 is the initial velocity, and s_0 is the initial position. These work for when the acceleration is zero, as in the horizontal component of velocity
s(t)_horz = (v_0_horz)t + (s_0_horz)
and
s(t)_vert = (1/2)at^2 + (v_0_vert)t + (s_0_vert)
Since projectile motion is divided in X and Y components, does that mean I can swap "x" for "y" in equations when I'm solving for the Y part...
NO
The x direction is horizontal. Horizontal motion is not affected by gravity, so the x velocity remains constant, unless air resistance is included in the problem.
The y direction is vertical, so the y velocity decreases 9.8 m/s each second as the object rises, and increases as the object falls.
does Vfy^2 = Viy + 2ayy?
The final vertical (y) velocity = Initial vertical velocity * time – (½ * g * time^2)
Initial vertical velocity = v * sin θ
Final vertical velocity = v * sin θ – g * t
Final vertical position = Initial vertical position + (v * sin θ * t) – (½ * g * t^2)
Initial horizontal velocity = v * cos θ
Final Horizontal position = v * cos θ * t
Maximum height occurs when vertical velocity = 0
Distance = average velocity * time
Time when vertical velocity = 0 is (v * sin θ) ÷ g
average velocity ½ * (v * sin θ)
Maximum height = ½ * (v * sin θ) * (v * sin θ) ÷ g
Maximum height = (v * sin θ)^2 ÷ (2 * g)
Nope, you can't swap. That results because the magnitude of the components varies as the angle of the direction of the velocity V. For example, Vx = V cos(theta) and Vy = V sin(theta). Clearly, unless theta = 45 deg, Vx <> Vy in general. At the extreme to illustrate. Suppose you fire straight up; so theta = 90 deg. Then Vx = V cos(90) = 0 and Vy = V sin(90) = V.
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