12.4: Shape of the Projectile Path
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What is the shape of the projectile's path in Fig. 12.1.1? To find out, we can solve Eq. (12.1.7) for the time t and plug the resulting expression into Eq. (12.1.8) to eliminate t and get an equation for y(x). First solve Eq. (12.1.7) for t :
t=xv0cosθ
Now substitute this into Eq. (12.1.8):
y=−12g(xv0cosθ)2+(v0sinθ)(xv0cosθ)
y(x)=(−g2v20cos2θ)x2+(tanθ)x
This is the equation of a parabola passing through the origin, so the projectile follows a parabolic path.
Actually, this is just an approximation, assuming the acceleration due to gravity is a constant downward in the −y direction. A more careful calculation would have to allow for the curvature of the Earth, which would show the actual path is that of a highly eccentric ellipse. But over relatively short distances where the curvature of the Earth is not important, the elliptical path can be approximated as a parabola.