A baseball's movement when considering air drag

[dan1el]

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

m \cdot \frac{\mathrm{d}v}{\mathrm{d}t} = -mg-Cv^2

\frac{\mathrm{d}v}{v^2 + \frac{mg}{C}} = -\frac{C}{m}\mathrm{d}t

-\frac{C}{m}t = \sqrt{\frac{C}{mg}}\arctan(\sqrt{\frac{C}{mg}}v) + A

V(t) = \sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t - B)

V(0) = \sqrt{\frac{mg}{C}}\tan(-B) = 45 m/s \Rightarrow B = -0.94

V(t) = \sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t + 0.94)

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:

\sqrt{\frac{mg}{C}}\tan(-\sqrt{\frac{Cg}{m}}t + 0.94) = 0

\sqrt{\frac{Cg}{m}}t = -B \Rightarrow t = -B \cdot \sqrt{\frac{m}{Cg}} = 3.17 s

Maple gives me the same answer.

What am I doing wrong?

 

[kuruman]

What you are doing wrong is that you are evaluating the indefinite integrals first, then try to apply the initial conditions. The correct way to do it is as shown below. Then v₀ will find its proper place.

<br /> \int^{v}_{v_{0}}\frac{\mathrm{d}u}{u^2 + \frac{mg}{C}} = -\int^{t}_{0}\frac{C}{m}\mathrm{d}t&#039;<br />

 

[dan1el]

I still get the same result.

\int^v_{v_0} \frac{\mathrm{d}u}{u^2 + \frac{mg}{C}} = \sqrt {{\frac {C}{mg}}} \left( \arctan \left( \sqrt {{\frac {C}{mg}}}v<br /> _{{0}} \right) -\arctan \left( \sqrt {{\frac {C}{mg}}}v \right) <br /> \right)

If you then let
B = -\arctan (\sqrt{{\frac {C}{mg}}}v_{{0}}),
you get what I had to begin with.

 

[kuruman]

You are right. If it is any consolation, I checked the numbers according to the last result above and my answer is 3.16 s. There may be a typo in the book, but I think that your analysis of this problem is correct and that there is very little left to learn from it.

 

[CFDFEAGURU]

I have the same result of 3.16 seconds.

 

[kuruman]

I have the same result of 3.16 seconds. I also have the result of 21.78 seconds to return to the Earth and a total height of 191.2 meters.

If by "total" you mean "maximum" height, then I think the number is too large. Without air resistance the maximum height v₀²/2g = 103.3 m.

 

[CFDFEAGURU]

Since air resistance is taken into account in the calculation of the time of 3.16 seconds how could the distance be incorrect?

It would seem that you could use the standard equation

x(t) = x(0) + V(0) * t + 1/2*a*t^2

to determine the maximum height. Otherwise, it would seem you are "doubling up" on the air resistance.

Correct me if I am wrong.

Thanks
Matt

 

[kuruman]

The kinematic equation is valid only if the acceleration is constant. In this problem it is not. It is velocity-dependent and given by

a = -g - Cv²

One needs to integrate the expression for v(t) to get x(t) then evaluate this last result at the the time of 3.16 s. If I did this correctly, the maximum height reached is 58.4 m. To calculate the time the ball takes to come down, one needs to solve the diff. eq. again because the resistive force changes sign relative to gravity.

 

[CFDFEAGURU]

Yes,

I agree with the above. I starting thinking the same as I drove into work this morning.

Thanks
Matt

 

[dan1el]

The answers in the book are wrong, apparently, so there's no problem after all. Thanks anyway!

 

[kuruman]

The consolation is that you know how to do the problem. It has nothing left to teach you.