How Can the Rocketeer Safely Catch the Falling Student from the CN Tower?

[Ali Asadullah]

A real challenge to all physicsist!

In order to experience free fall a student walks off the top of the CN Tower which is 553m high and falls freely. His initial velocity is zero. The rocketeer arrives at the scene 5 seconds 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 get 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 accomplish this is provided by the rocketeer's jet pack. which is turned on just as he catch the student. Before then the rocketeer is in free fall. To prevent discomfort to the student the magnitude of the acceleration of rocketeer and the student as they move downward together should be no more that 5 times g. What is the minimum height above the ground at which the rocketeer should catch the student? What must be the rocketeer's initial downward speed so that he catch the student at the minimum height found above.

 

[tiny-tim]

Hi Ali!

Show us what you've tried, and where you're stuck, and then we'll know how to help!