- #1
dan1el
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Homework Statement
A baseball, with mass m = 0.145 kg, is thrown directly upward from z(0) = 0 m, with initial speed v(0) = 45 m/s. The air drag on the ball is given by Cv2, C = 0.0013 Ns2/m2. Set up a diff. eq. for the ball's movement and solve it for both v(t) and z(t).
m = 0.145 kg
C = 0.0013 Ns2/m2
g = 9.81 m/s2
v(0) = 45 m/s
z(0) = 0 m
Homework Equations
F = ma
The Attempt at a Solution
[tex]m \cdot \frac{\mathrm{d}v}{\mathrm{d}t} = -mg-Cv^2[/tex]
[tex]\frac{\mathrm{d}v}{v^2 + \frac{mg}{C}} = -\frac{C}{m}\mathrm{d}t[/tex]
[tex]-\frac{C}{m}t = \sqrt{\frac{C}{mg}}\arctan(\sqrt{\frac{C}{mg}}v) + A[/tex]
[tex]V(t) = \sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t - B)[/tex]
[tex]V(0) = \sqrt{\frac{mg}{C}}\tan(-B) = 45 m/s \Rightarrow B = -0.94[/tex]
[tex]V(t) = \sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t + 0.94)[/tex]
The ball is supposed to reach its maximum height after 3.36 s, according to the book. However, when I set v(t) = 0, I get this:
[tex]\sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t + 0.94) = 0[/tex]
[tex]\sqrt{\frac{Cg}{m}}t = -B \Rightarrow t = -B \cdot \sqrt{\frac{m}{Cg}} = 3.17 s[/tex]
Maple gives me the same answer.
What am I doing wrong?