Baseball Projectile Motion Problem

  • #1
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 [tex]\overrightarrow{v}[/tex] and the horizontal be so that the ball crosses the plate 0.7 m above the ground?

The following values are given:

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

The following should also be noted:

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

The equation of motion in the x direction is:

[tex]x = x_0 + v_{0x}t[/tex]
[tex]\Rightarrow x = v_{0x}t[/tex]
[tex]\Rightarrow t = \frac{x}{v_{0x}}[/tex]
[tex]\Rightarrow t = \frac{x}{v_0 \cos \theta_0}[/tex]
[tex]\Rightarrow t = \frac{18.4}{37.5 \cos \theta_0}[/tex]

The equation of motion in the y direction is:

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

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

[tex]-1.1809 + 9.2 \sin(2 \theta_0) + 1.8 (1 - \sin^2 \theta_0) = 0[/tex]
[tex]\Rightarrow -1.1809 + 9.2 \sin(2 \theta_0) + 1.8 - 1.8 \sin^2 \theta_0 = 0[/tex]
[tex]\Rightarrow -1.8 \sin^2 \theta_0 + 9.2 \sin(2 \theta_0) + 0.6191 = 0[/tex]

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

The answer given in the back of the book is [itex]\theta_0 = -1.93^\circ[/itex].

Note also that:

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

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?
 

Answers and Replies

  • #2
56
0
as a suggestion, change [tex]\sin{2\theta_0}=2\sin{\theta_0}\cos{\theta_0}[/tex]
wisky40
 
  • #3
56
0
and also [tex]\cos^2{\theta}+\sin^2{\theta}=1
 
  • #4
56
0
once again [tex]\cos^2{\theta}+\sin^2{\theta}=1[/tex]
 
  • #5
Doc Al
Mentor
45,248
1,598
NoPhysicsGenius said:
The equation of motion in the x direction is:

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

The equation of motion in the y direction is:

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

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 [itex]\theta[/itex].
 
  • #6
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. :biggrin:

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 [itex]\theta[/itex].

Nice tip!

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

[tex]24.059t^4 - 1423.908t^2 +341.8 = 0[/tex]

Applying the quadratic formula gave two values for [itex]t^2[/itex] which, after taking the square root, gave two positive values for [itex]t[/itex]. They were [itex]t = 0.490945s[/itex] and [tex]t = 7.67743s[/itex].

Naturally, since the pitcher throws a fast ball, we should use [tex]t = 0.490945s[/tex].

Then:

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

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

Therefore, the book's answer of [itex]\theta_0 = -1.93^\circ[/itex] 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:
 

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