# Baseball Projectile Motion Problem

I am having difficulty with Problem 79 of Chapter 3 from Physics for Scientists and Engineers by Paul A. Tipler, 4th Edition:

The distance from the pitcher's mound to home plate is 18.4 m. The mound is 0.2 m above the level of the field. A pitcher throws a fast ball with an initial speed of 37.5 m/s. At the moment the ball leaves the pitcher's hand, it is 2.3 m above the mound. What should the angle between $$\overrightarrow{v}$$ and the horizontal be so that the ball crosses the plate 0.7 m above the ground?

The following values are given:

$$y_0 = 0.2 m + 2.3 m = 2.5m$$
$$y = 0.7 m$$
$$x_0 = 0$$
$$x = 18.4 m$$
$$v_0 = 37.5 m/s$$

The following should also be noted:

$$v_{0x} = v_0 \cos \theta_0$$
$$v_{0y} = v_0 \sin \theta_0$$
$$g = 9.81 m/s^2$$

The equation of motion in the x direction is:

$$x = x_0 + v_{0x}t$$
$$\Rightarrow x = v_{0x}t$$
$$\Rightarrow t = \frac{x}{v_{0x}}$$
$$\Rightarrow t = \frac{x}{v_0 \cos \theta_0}$$
$$\Rightarrow t = \frac{18.4}{37.5 \cos \theta_0}$$

The equation of motion in the y direction is:

$$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$$
$$\Rightarrow 0.7 = 2.5 + (v_0 \sin \theta_0)t - 4.905t^2$$
$$\Rightarrow -4.905 (\frac{18.4}{37.5 \cos \theta_0})^2 + (37.5 \sin \theta_0)(\frac{18.4}{37.5 \cos \theta_0}) + 1.8 = 0$$
$$\Rightarrow -1.1809 \frac{1}{cos^2 \theta_0} + 18.4 \frac{\sin \theta_0}{\cos \theta_0} + 1.8 = 0$$
$$\Rightarrow -1.1809 + 18.4 \sin \theta_0 \cos \theta_0 + 1.8 \cos^2 \theta_0 = 0$$

Noting that $\cos^2 \theta_0 + \sin^2 \theta_0 = 1 \Rightarrow \cos^2 \theta_0 = 1 - \sin^2 \theta_0$, and also that $\sin(2 \theta_0) = 2 \sin \theta_0 \cos \theta_0$, we then have:

$$-1.1809 + 9.2 \sin(2 \theta_0) + 1.8 (1 - \sin^2 \theta_0) = 0$$
$$\Rightarrow -1.1809 + 9.2 \sin(2 \theta_0) + 1.8 - 1.8 \sin^2 \theta_0 = 0$$
$$\Rightarrow -1.8 \sin^2 \theta_0 + 9.2 \sin(2 \theta_0) + 0.6191 = 0$$

How do I solve for $\theta_0$ without resorting to graphical methods (this is an intro physics text, after all)?

The answer given in the back of the book is $\theta_0 = -1.93^\circ$.

Note also that:

$$-1.8 \sin^2 (-1.93^\circ) + 9.2 \sin[(2)(-1.93^\circ)] + 0.6191 = -0.00227... \approx 0$$

It therefore appears that I have set the problem up correctly; I just don't know how to put the finishing touch on the solution. Any ideas?

as a suggestion, change $$\sin{2\theta_0}=2\sin{\theta_0}\cos{\theta_0}$$
wisky40

and also $$\cos^2{\theta}+\sin^2{\theta}=1 once again [tex]\cos^2{\theta}+\sin^2{\theta}=1$$

Doc Al
Mentor
NoPhysicsGenius said:
The equation of motion in the x direction is:

$$x = x_0 + v_{0x}t$$
$$\Rightarrow x = v_{0x}t$$
$$\Rightarrow t = \frac{x}{v_{0x}}$$
$$\Rightarrow t = \frac{x}{v_0 \cos \theta_0}$$
$$\Rightarrow t = \frac{18.4}{37.5 \cos \theta_0}$$
Looks good to me. But don't be in such a hurry to solve for t. The equation can also be written like this:
$$18.4 = 37.5 cos\theta t$$

The equation of motion in the y direction is:

$$y = y_0 + v_{0y}t - \frac{1}{2}gt^2$$
$$\Rightarrow 0.7 = 2.5 + (v_0 \sin \theta_0)t - 4.905t^2$$
$$\Rightarrow -4.905 (\frac{18.4}{37.5 \cos \theta_0})^2 + (37.5 \sin \theta_0)(\frac{18.4}{37.5 \cos \theta_0}) + 1.8 = 0$$
$$\Rightarrow -1.1809 \frac{1}{cos^2 \theta_0} + 18.4 \frac{\sin \theta_0}{\cos \theta_0} + 1.8 = 0$$
$$\Rightarrow -1.1809 + 18.4 \sin \theta_0 \cos \theta_0 + 1.8 \cos^2 \theta_0 = 0$$
Once again, don't rush. Think strategically. In order to use the $sin^2\theta + cos^2\theta = 1$ trick, you need to isolate those trig functions. The equation for y can also be rewritten like this:
$$4.9t^2 - 1.8 = 37.5 sin\theta t$$

Now do you see how to apply that trig identity? (Square both equations and add them together.) Solve for t, then use t to solve for $\theta$.

Doc Al said:
Once again, don't rush. Think strategically.

Hmm ... I don't believe I've ever tried this before. Might also explain my difficulties with winning games of chess.

Doc Al said:
Now do you see how to apply that trig identity? (Square both equations and add them together.) Solve for t, then use t to solve for $\theta$.

Nice tip!

After squaring both equations and adding, I got the following:

$$24.059t^4 - 1423.908t^2 +341.8 = 0$$

Applying the quadratic formula gave two values for $t^2$ which, after taking the square root, gave two positive values for $t$. They were $t = 0.490945s$ and $$t = 7.67743s[/itex]. Naturally, since the pitcher throws a fast ball, we should use [tex]t = 0.490945s$$.

Then:

$$18.4 = 37.5(\cos \theta_0)t \Rightarrow \theta_0 = cos^{-1}\frac{18.4}{37.5t}$$
$$\Rightarrow \theta_0 = cos^{-1}\frac{18.4}{37.5(0.490945)} = \pm 1.93^\circ$$

To determine whether $\theta_0$ is positive or negative, one must then plug the solution back into the expression for $y$ (namely, $y = y_0 + v_{0y}t - \frac{1}{2}gt^2$), which yields $y = 1.94 m \neq 0.7m$ for $\theta_0 = +1.93^\circ$ versus $y = 0.7 m$ for $\theta_0 = -1.93^\circ$.

Therefore, the book's answer of $\theta_0 = -1.93^\circ$ is vindicated as correct.

Thanks for your help. I believe my approach to solving physics problems in the future will improve as a result.