I am having difficulty with Problem 79 of Chapter 3 from Physics for Scientists and Engineers by Paul A. Tipler, 4th Edition:(adsbygoogle = window.adsbygoogle || []).push({});

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?

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