# Initial Position Discrepancy Involving Fluid Resistance

• TRB8985
In summary, the rock in the simple projectile motion case starts from a lower position and experiences a greater drag than in the case with fluid resistance proportional to v.

#### TRB8985

Homework Statement
A small rock moves in water and the force exerted on it by the water is given by f = kv.

The terminal speed of the rock is 2.00 m/s, and the rock is projected upward at an initial speed of 6.00 m/s. Ignore the bouyancy force on the rock.

a.) Without fluid resistance, how high will the rock rise & how long does it take to reach this maximum height?

b.) When fluid resistance is included, what are the answers to the question in part (a)?
Relevant Equations
1.) f = kv

2.) v_y(t) = v_t * (1 - exp(-kt/m)) + v_0(exp(-kt/m))

3.) y(t) = v_t*t + m/k*exp(-kt/m)*(v_t - v_0)
Good evening,

I have a question on the problem I've provided in the homework statement. Essentially, the problem is asking us to compare the maximum height and time to max-height between a simple projectile motion case vs. a case using fluid resistance proportional to v. I've solved everything correctly and my answers match the solution in the back of my book, but there's something odd I can't seem to wrap my head around.

In the simple projectile motion case with no fluid resistance, the rock starts at y = 0 and climbs to a maximum height of 1.84m in 0.612 seconds using basic kinematics equations. No problem there.

When fluid resistance is involved, I integrated the 2nd equation to get the 3rd, using t = 0 to the time of max-height of 0.283s. The details involving that process aren't too important here since it's correct, but this is what I found a little confusing:

At t = 0 using equation 3, I get an initial starting point of y = 1.63m. (Positive, because in part B I set the downward direction to +y)

Is there a reason that the rock is displaced 1.63m away from y = 0 when fluid resistance is included at the beginning time? Seems a little strange that the rocks wouldn't both start at y = 0. Instead, when fluid resistance turns on, the rock starts from a much lower position despite not having any time to move compared to the basic projectile motion version.

Hopefully that makes sense. Thank you for your input.

TRB8985 said:
I integrated the 2nd equation to get the 3rd
… but forgot to allow an integration constant.

• TRB8985
haruspex said:
… but forgot to allow an integration constant.
Oh! That actually makes sense. Looks like our book answer was missing that portion.

Thank you!

TRB8985 said:
Oh! That actually makes sense. Looks like our book answer was missing that portion.

Thank you!
Btw, it is unclear how you got the 2nd equation. You cannot in general treat horizontal and vertical independently in drag calculations, but it happens to work out ok in the case of linear drag. I'd like to see exactly what you did.

Sure, be happy to share if you're interested in seeing. As far as I'm aware, this should be a fluid resistance problem that's entirely in the vertical direction.

Regarding the 2nd equation, that actually came from an earlier problem where we had to derive an expression for a falling object with an initial downward velocity of v_0 experiencing the same kind of drag proportional to v.

It's a bit of work to write all out, but here's the jist:

1.) Started with the FBD in the y-direction (+y downward). Had the following: W - kv = ma

2.) Exchanged a with dv/dt (or rather, dv/dt' since I was integrating from 0 to t and didn't want the integration variable to be the same as the variable in the integration limit).

3.) Exchanged W with mg, and divided both sides by k.

4.) Replaced mg/k with a variable to represent the terminal velocity, v_t.

5.) Isolated dv/dt' on the RHS and used separation of variables, resulting in the following:

k/m * dt' = dv / (v_t - t)

6.) Integrated the LHS from 0 to t, and the RHS from v_0 to the final velocity in the y-direction, v_y.

7.) Resulted in the following after a couple steps of algebra:

-kt/m = ln( v_t - v_y / v_t - v_0 )

8.) Took the exponential of both sides to crack open the natural logarithm and multiplied both sides by (v_t - v_0).

9.) Couple steps of algebra later, got the following:

v_y = v_t - v_t * exp(-kt/m) + v_0*exp(-kt/m)

10.) Factored the v_t terms together to resulting in equation #2. This also matched our book's answer, so I felt pretty confident about that process.

Hopefully that wasn't too brief!

TRB8985 said:
Sure, be happy to share if you're interested in seeing. As far as I'm aware, this should be a fluid resistance problem that's entirely in the vertical direction.
Sorry, for some reason I thought the water was flowing, making for a horizontal motion too. But thanks for posting anyway. It makes the thread more useful to others.