Nick hits a baseball 3 feet above the ground at an angle of $25^\circ$ and a speed of 150 ft/sec straight toward center field. At the moment the ball is hit, there is a 5mph (8.8 ft/sec) wind blowing straight back in from center field.

I'm supposed to write two parametric equations that simulate the path of the baseball with respect to time.

So I've tried solving this problem by setting up a right triangle with angle $25°$ and hypotenuse $150 - 8.8 = 141.2$ because though the speed is $150 ft/sec$, the wind blows from the other side at $8.8 ft/sec$... However, in the answer key they seem to have not noted this $8.8 ft/sec$ at all and simply put $150$ as the hypotenuse. But they provide no explanation, so can anyone tell me why it is so? Why don't we take the speed of the wind into account?

  • $\begingroup$ What physical law do you have in hand, to deal with wind resistance? Are you to use soemthing such as $F = ma$, where $F= \alpha v$ represents the effect of resistance ($\alpha$ being a coefficient)? Adding, in whatever way, the wind speed to the initial speed seems to suggest other approaches. $\endgroup$ – An aedonist Aug 7 '17 at 12:18

Here's what I would do. (Note: ignoring air resistance). It might also be useful to look at the kinematic equations of motion if you're interested in the physics.

Anyway, it seems you want $p(t)=(x(t),y(t))$ to represent the ball's position via parametric equations. The equations are 2D because we can look at the plane going between the hitter and pitcher (no forces push the ball out of the plane).

We can first decompose the initial velocity into $x$ and $y$: let $\theta=25^\circ$, then $x'(0)=150\cos(\theta)=v_x$ and $y'(0)=150\sin(\theta)=v_y$, with $x(0)=0$ and $y(0)=y_0=3$.

So in $x$, the only change in velocity is the wind $w=-8.8$; thus, $$ x'(t) = v_x + wt \;\;\;\implies\;\;\; x(t)=v_xt + \frac{1}{2}wt^2$$ note that we've assumed the wind exactly imparts its velocity i.e. $x(0)=0$ and $x(\tau)=0$, where $\tau = -2v_x/w$.

For $y$, gravity pushes down all the time with force $f=mg$ i.e. $y''(t)=g$, so $y'(t)=gt+C_1$ and since $y'(0)=v_y$, we get $C_1 = v_y$. Thus, $y(t)=gt^2/2+v_yt+C_2$, so the initial conditions give:$$ y(t)=\frac{1}{2}gt^2+v_yt+y_0$$

So the final parametric equations from this approach are: $$ p(t)=\begin{bmatrix} v_x\\ v_y \end{bmatrix}t + \frac{1}{2} \begin{bmatrix} w\\ g \end{bmatrix}t^2 + \begin{bmatrix} 0\\ y_0 \end{bmatrix} $$

If anyone sees a mistake let me know!


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