Question

Use the model for projectile motion, assuming there is no air resistance. Eliminate the parameter \(t\) from the position function for the motion of a projectile to show that the rectangular equation is \(y=-\frac{16 \sec ^{2} \theta}{v_{0}^{2}} x^{2}+(\tan \theta) x+h\)

Short answer

The above solution steps have shown that by manipulating the physics equations of motion, it is possible to derive the given equation for a projectile's parabolic trajectory with the horizontal and vertical distances as \(x\) and \(y\), respectively, and neglecting air resistance. This equation typically refers to a parabola that opens downward with the parameter \(h\), indicating the vertical shift, as 0 in this case.

Step-by-step solution

Extract \(t\) from the \(x\)-equation

First, start with the horizontal motion equation \(x = v_0 \cos(\theta) t\) and solve for \(t\), which gives \(t= \frac{x}{v_0 \cos (\theta)}\). This will be useful for replacing in the \(y\)-equation.

Substitute \(t\) into \(y\)-equation

Next, substitute \(t\) from the first step into the equation for \(y\), which is \(y = v_0 \sin(\theta) t - 16t^2\). This gives \(y = v_0 \sin(\theta) \frac{x}{v_0 \cos (\theta)} - 16 \left(\frac{x}{v_0 \cos (\theta)}\right)^2\). It simplifies to \(y= x \tan(\theta) - \frac{16x^2}{v_0 ^2 \cos^2(\theta)}\). You can observe that \(\sec^2(\theta)\) is the reciprocal of \(\cos^2({\theta})\) so that you can make the substitution to get closer to the given formula.

Apply trigonometric identity

Use the identity \(\sec^2(\theta) = 1 / \cos^2(\theta)\) to replace \(\cos^2(\theta)\) in above equation. This yields \(y= x \tan(\theta) - 16 \sec^2(\theta) \frac{x^2}{v_0 ^2}\).

Re-arrange the equation

Finally, it is a matter of rearranging terms to match the given equation, which yields \(y=-\frac{16 \sec ^{2} \theta}{v_{0}^{2}} x^{2}+(\tan \theta) x + 0\). The \(+h\) term in the given equation indicates a vertical shift in the parabola, but as it was not initially presented in our original equation, is understood here to be 0.