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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 !!!

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# Homework Help: Free Fall Rocket Acceleration Problem

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