# Free Fall Acceleration Problem

1. Jan 21, 2007

### peoplelikeus

1. The problem statement, all variables and given/known data

A student walks off the top of the CN Tower in Toronto, which has height 553m and falls freely. His initial velocity is zero. The Rocketeer arrives at the scene a time of 4.75 seconds later and dives off the top of the tower to save the student. The Rocketeer leaves the roof with an initial downward speed of v_0.

In order both to catch the student and to prevent injury to him, the Rocketeer should catch the student at a sufficiently great height above ground so that the Rocketeer and the student slow down and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the Rocketeer's jet pack, which he turns on just as he catches the student; before then the Rocketeer is in free fall. To prevent discomfort to the student, the magnitude of the acceleration of the Rocketeer and the student as they move downward together should be no more than five times g.

PART A. What is the minimum height above the ground at which the Rocketeer should catch the student?

PART B. What must the Rocketeer's initial downward speed be so that he catches the student at the minimum height found in part (a)? Take the free fall acceleration to be g = 9.8m/s^2.

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I've tried solving this for a long time and both Part A and Part B are really difficult. Here are the equations I have come up with...

Student's Position; s(t) = 553-4.9t^2
Rocketeer's Position; r(t) = 553-v_0(t+4.75)-4.9(t+4.75)^2

For part A, I tried calculating the height with this equation:

0 = height caught - student's velocity when caught + .5(g)t^2.

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Beyond this, I am completely lost ! :( Any help would be appreciated !!!

2. Jan 21, 2007

### PhanthomJay

When the rocketeer grabs the student at some point 'd' above the ground, they both have the same initial speed at that point. And they both have the same final speed of 0 when they hit the ground. And the upward acceleration is given as 5g's. That should give you one equation with 2 unknowns, V and d. What's the other? For the second part, when solving for the rocketeer's initial speed, note that the rocketeer catches up with the student in (T-4.5) seconds from his start, where T is the time measured by the student from her start. I believe in your equations you slipped on the plus and minus sign.

3. Jan 21, 2007

### peoplelikeus

For the part A, are you suggesting that the first equation I need is,

0 = d - vt + .5(5g)t^2

erm well, I don't think that's right since I have three unknowns. I don't think I understood your wording.

4. Jan 22, 2007

### PhanthomJay

Leave time out of it for the moment, and try
$$v_f^2 = v_o^2 - 2ad$$ where a = 5g. There are only 2 unknowns in this one.

5. Feb 7, 2007

### dimeking

For your reference, this problem is a slightly modified version of Problem 2.91 on page 77 from Young and Freedman’s 11th Edition of University Physics. In Y&F’s version, the Rocketeer leaves the building roof 5.00 seconds after the student. The answers from the back of the book are h = 92.2m and v0 = 75.1m/s.

I agree with PhantomJay that your equation for the Rocketeer’s position should include the term t–4.75 and not 4+4.75.

Beyond this, I think the key to solving this is to determine the velocity of the Rocketeer/student just after the Rocketeer catches the student. Just before the Rocketeer catches the student, the magnitude of his velocity is greater than that of the student. Conservation of linear momentum is not introduced until Chapter 8 of Y&F’s book. Including momentum and assuming that the Rocketeer/student’s acceleration can be changed instantaneously by the jet pack and that air resistance may be ignored, the velocity of the 2 of them will be something in between the velocities they each had just before the catch.

Therefore, I believe the real (and unsatisfying) answer is that there is not enough information given to solve the problem.

I ignored this (trying to get the answers in the book) and set up equations for the student and Rocketeer just before the catch.

Starting with y = y0 +(v0)t +(1/2)a(t^2), I got h = 553 – 4.9(t1)^2 for the student and h = 430.5 – (t1)(v0) + 5(v0) + 49(t1) – 4.9(t1)^2 for the Rocketeer, where t1 is the time of the catch and h is the height above ground at the time of the catch. Keep in mind, I used 5.00 seconds between the times that the student and Rocketeer leave the roof.

I combined these 2 equations, eliminating t1. This gave me a somewhat messy equation in terms of h and v0, the desired unknowns. Inserting the book’s answers showed this to agree with Y&F so far.

I then set up equations for the Rocketeer/student just after the catch. First, I assumed the velocity of the 2 of them equals the velocity of the Rocketeer just before the catch. I used these equations and tried to combine them in many ways with the equations for motion just before the catch. I tried this more times than I care to admit and came up with nonsense answers, trivial answers, and 4th degree equations that I couldn’t solve with simple algebra.

Giving up that approach, I assumed the velocity of the 2 of them equals the velocity of the student just before the catch. I used these equations, combined them with the equations for motion just before the catch, and with relative ease got the answers that agree with Y&F.