Baseball Projectile Motion Problem

[NoPhysicsGenius]

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?

[wisky40]

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

[wisky40]

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

[wisky40]

once again \cos^2{\theta}+\sin^2{\theta}=1

[Doc Al]

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.

[NoPhysicsGenius]

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. :biggrin:

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. :cool: