How to extract Parametric Equations from a Problem

Question

How can I extract parametric equations from a scenario which is given in a problem like this:

A hitter's bat meets a ball about 3 ft above home plate. Suppose the horizontal component of the ball's initial velocity is 100 ft/s and the vertical component is 60 ft/s.

The equations that was shown in example were:

\begin{align*} x&= 100t\\ y&=3+60t-16t^2 \end{align*}

Answer 1

You need to understand what are the equations of motion for the ball. You can decompose the motion into two components - horizontal and vertical. In the horizontal direction you have no acceleration, so the velocity is constant (equal to the initial horizontal velocity). If you assume the initial $x$ coordinate to be $0$, $$x=x_0+v_xt$$ becomes $$x=0+100t=100t$$ In the vertical direction you start at $y_0=3$, with a vertical velocity upwards $v_y=60$. The acceleration is downwards $g=-32ft/s^2$. The equation of motion for constant acceleration is $$y=y_0+v_yt+\frac12gt^2=3+60t-16t^2$$

Answer 2

Let's go through this line-by-line, and translate the written statements into mathematical statements. To begin, let $x(t)$ and $y(t)$ be the horizontal and vertical positions, in feet, of the ball in relation to home plate at time $t$. Then, we have the following:

A hitter's bat meets a ball about $3$ ft above home plate.

This tells us that that at time $t = 0$ the ball is at $(x,y) = (0,3)$, because it is at home plate ($x=0$), but $3$ ft off the ground ($y=3$). This gives us $x(0) = 0$, and $y(0) = 3$.

The horizontal component of the ball's initial velocity is $100$ ft/s

This tells us that the $x'(0) = 100$, because $x'(t)$ is the horizontal component of the ball's velocity.

The vertical component (of the ball's initial velocity) is $60$ ft/s.

As before, this tells us that $y'(0) = 60$, because $y'(t)$ is the vertical component of the ball's velocity.

So, we've established the following information: \begin{align*} x(0) &=0\\ y(0) &= 3\\ x'(0) &= 100\\ y'(0) &= 60. \end{align*}

Lastly, we need to remember an additional assumption about projectile motion, namely that acceleration is only due to gravity. Remembering that $x''(t)$ and $y''(t)$ are the horizontal and vertical components of acceleration, respectively, this provides us with \begin{align*} x''(t) &= 0 \text{ ft/s$^2$}\\ y''(t) &= -32 \text{ ft/s$^2$}. \end{align*}

So, working our way backwards, first for $x(t)$: \begin{align*} x''(t) = 0 &\implies x'(t) = c_1\\[5pt] x'(0) = 100 &\implies c_1 = 100 \implies x'(t) = 100\\[5pt] x'(t) = 100 &\implies x(t) = 100t + c_2\\[5pt] x(0) = 0 &\implies c_2 = 0 \implies \boxed{x(t) = 100t} \end{align*}

Next, for $y(t)$: \begin{align*} y''(t) = -32 &\implies y'(t) = -32t + c_3\\[5pt] y'(0) = 60 &\implies c_3 = 60 \implies y'(t) = -32t + 60\\[5pt] y'(t) = -32t + 60 &\implies y(t) = -16t^{2} + 60t + c_4\\[5pt] y(0) = 3 &\implies c_4 =3 \implies \boxed{y(t) = -16t^2 + 60t + 3} \end{align*}