Determined to test gravity a student walks off the CN Tower in Toronto, which is

Question

Determined to test gravity a student walks off the CN Tower in Toronto, which is 553m high and falls freely. His initial velocity is zero. The Rocketeer leaves the roof with an initial velocity downward and then is in freefall. In order to both catch the student and to prevent injury to him, the Rocketeer should catch the student at a sifficiently great height and arrive at the ground with zero velocity. The upward acceleration that accomplishes this is provided by the Rocketter's jet pack, which he turns on just as he catches the student; before then the student is in freefall. To prevent discomfort to the student the magnitude of the acceleration is limited to fives times gravity. How high above the ground must the Rocketeer catch the student?

*Thanks in advance for the much needed help!

Explanation / Answer

let the distance above ground at which the rocketeer catches the student be Xm.
so velocity of the student at the point when he is caught by the rocketeer is
V12-0=2g(553-X) {as v2 - u2 = 2as}
when they reach ground the velocity(V2) ceases to zero,this implies
V22 - V12 = 2(-5g)X

=> 0 - 2g(553-X) = -5gX

=> 553-X=2.5X

=> 3.5X=553

X=158m

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