WebAssign Homework - Basic Acceleration Problem

[CoryG89]

First off, this is my first post, so hi everybody, this looks like a good place I can get some real help possibly. I am taking a Physics I with Calc course at my college and I am having to use an internet based homework service that comes with our textbook called WebAssign. This seems to be hindering me at my work a little so be gentle if the solution is blindingly obvious. Thank you.

Homework Statement

Every morning at seven o'clock
There's twenty terriers drilling on the rock.
The boss comes around and he says, "Keep still
And bear down heavy on the cast-iron drill
And drill, ye terriers, drill." And drill, ye terriers, drill.
It's work all day for sugar in your tea
Down beyond the railway. And drill, ye terriers, drill.

The foreman's name was John McAnn.
By God, he was a blamed mean man.
One day a premature blast went off
And a mile in the air went big Jim Goff. And drill...
Then when next payday came around
Jim Goff a dollar short was found.
When he asked what for, came this reply:
"You were docked for time you were up in the sky."
And drill...
-- American folksong

If Goff actually went 1.00 mile in the air and was docked exactly $0.95, what was Goff's hourly wage?

Homework Equations

We are given several kinematic equations to work with in this chapter. I am assuming this falls under free falling motion with constant acceleration due to gravity.

I was using this equation: x_f = x_i + v_xit + (1/2)a_xt²

The Attempt at a Solution

I am not sure that I signed up for poetic interpretation with my physics course, but apparently they expect me to be creative aswell. The only way I can interpret this problem is saying that he was shot up (apparently almost instantly) 1 mile into the air. At this point when he obtains his maximum height his velocity should be 0. From this point on he falls due to the constant acceleration of gravity which in this case would be -9.80.

In order to obtain the time that it took for him to reach the ground we need to substitute the known information into the equation (I convert 1 mile into 1609 meters using SI units) and solve for t. Doing this I get:

x_f = x_i + v_xit + (1/2)a_xt²
0 = 1609 + 0t + (1/2)(-9.80)t²

Solving for t I obtain:

t = 18.1209 seconds.

From this information. It would lead me to believe that Goff was docked $0.95 for being in the air for 18.1209 seconds. This would lead to a huge hourly wage.

So he was docked $0.95 for .005 hours. This would lead to a huge hourly wage and I know i am doing something wrong. I just do not understand. I do not see how any other equations will work with this problem and the given information. If anyone can help it would be much appreciated.

 

[inutard]

Hello Cory, your work seems mostly right in principle but you may want to check the time he was up in the air again. Perhaps you may want to consider the total time he was in the air (the time going up and the time going down). How would such a consideration change your calculations and the final answer?

Keep working at it!

 

[Delphi51]

Welcome to PF!

You can't expect the poet to have worked out the time and got the pay correct!
Just go with the physics. Your calc is correct for the time to fall, but you have missed the time to go up. I think you can safely assume the explosion gives the guy his initial velocity in negligible time and distance, but it will than take a significant amount of time to go up to his maximum height.

 

[CoryG89]

This are true observations, but this is all the information I am given, and I cannot see how I am supposed to derive an initial velocity from the information. Perhaps the question is incomplete, although I doubt it.

 

[Delphi51]

Oh, I think the info is complete all right. Say you launched a toy rocket straight up and it went a mile high. You should be able to calculate the initial velocity and the time of flight just using y = Vi*t + ½at² and Vf = Vi + at.

You may also be able to leap to the answer by using symmetry. Sketch the graph of height vs time for the whole flight and mark in the time you have already calculated!