- #1
GZM
- 14
- 1
Homework Statement
Problem states : A student walks off the top of the CN Tower in Toronto, which has height h and falls freely. His initial velocity is zero. The Rocketeer arrives at the scene a time of t later and dives off the top of the tower to save the student. The Rocketeer leaves the roof with an initial downward speed v0. 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.
It asks for : What is the minimum height above the ground at which the Rocketeer should catch the student?
and :
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.)
Homework Equations
Kinematic equations :
d=vi*t+(1/2)*a*t^2
vf^2 = vi ^2 + 2*a*d
vf = vi + at
The Attempt at a Solution
My attempt of the solution was this diagram, but I don't know what I can equate other than d to each other to solve, but that won't give me anything for the distance that the Rocketeer catches the student, but all I know is the d is equivalent for both of these guys