# Drag coefficient question

Tags:
1. Sep 6, 2015

### SYoungblood

1. The problem statement, all variables and given/known data
A baseball is hit at a height of 3 ft off thte ground. It leaves the bat with an initial speed of 165ft/sec, at an angle of 10 degrees from the horizontal. The linear drag coefficient is k=0.38; gravitational acceleration is 32 ft/sec^2

2. Relevant equations

Find a vector form for the path of the baseball using these equations for linear drag,

x=V(sub)o/k (1- e^-kt) cos 10

y=(sub)o +V(sub)o/k (1- e^-kt) sin 10 + (g/k^2)(1- kt- e^-kt)

What is a vector form for the path of the baseball? Use only rational numbers in the expression as coefficients for the i and j vectors

3. The attempt at a solution

Oh my, I am lost in the sauce on this one…

x=165/0.38(1 - e^-0.38t) *cos 10 = 427.61389 (1 - e^-0.38t)

y= 3 + 165/0.38(1 - e^-0.38t)*sin10 + 32/0.1444(1 + 0.38t - e^0.38t) = 3 + 81.8627(1 - e^-0.38t)+ 221.6066(1 + 0.38t - e^0.38t)

Well, I combined the constants, but how to get rid of the t variable in the equation is something that I simply am not seeing, any and all help is appreciated.

Once I get the vector equation, then my problem asks for things I assume are fairly standard for this type of equation, time in flight, range of the hit ball, does it clear the fence z feet away. That is all fairly straightforward, but have mercy, I have never seen this drag coefficient before and this is proving a tough starting block.

SY

2. Sep 6, 2015

### SteamKing

Staff Emeritus
The i and j position vectors (or x and y, if you prefer) are parametric equations in t. I don't think, given the nature of these equations, you can eliminate the parameter t. Besides, what purpose would elimination of the parameter serve in the solution of this problem anyway?

3. Sep 6, 2015

### SYoungblood

That was a thought that crossed my mind, and the only purpose I can see was that it answered the question on my homework. All the followup problems stem from it.

I think I will just write this off as a loss and drive on.

Thanks just the same,

SY

Last edited: Sep 6, 2015