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 \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.

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].<br /> <br /> Naturally, since the pitcher throws a <b><i>fast ball</i></b>, we should use t = 0.490945s.<br /> <br /> Then:<br /> <br /> 18.4 = 37.5(\cos \theta_0)t \Rightarrow \theta_0 = cos^{-1}\frac{18.4}{37.5t}<br /> \Rightarrow \theta_0 = cos^{-1}\frac{18.4}{37.5(0.490945)} = \pm 1.93^\circ<br /> <br /> 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.<br /> <br /> Therefore, the book&#039;s answer of \theta_0 = -1.93^\circ is vindicated as correct.<br /> <br /> Thanks for your help. I believe my approach to solving physics problems in the future will improve as a result. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60e.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":cool:" title="Cool :cool:" data-smilie="6"data-shortname=":cool:" />