Q. A determined student walks off the top of the CN Tower inToronto, which is 55
ID: 1686165 • Letter: Q
Question
Q. A determined student walks off the top of the CN Tower inToronto, which is 553m high, and falls freely. His initial velocityis zero. The Rocketeer arrives at the scene 5s later and dives ofthe top of the tower to save the student. The rockereer leaves theroof with a downward speed Vo. In order to catch thestudent and to prevent injury to him the Rocketeer should catch thestudent at a sufficiently great height above ground so that theRocketeer and the student slow down and arrive at the ground withzero velocity. The upward acceleration that accomplishes this isprovided by the Rocketeer's jet pack, which he turns on just as hecatches the student; before then the Rocketeer is in free fall. Toprevent discomfort to the student, the magnitude of theacceleration of the Rocketeer and the student as they move downwardtogether should be no more than 5g.a) What is the minimum height above ground at which therocketeer should catch the student?
b) What must the Rocketeer's initial downward speed be sothat he catches the student at the minimum height found in part(a)?
Explanation / Answer
let the rocketeer catch the student at time ts distance covered by student in t sec. = 0*t + 0.5*9.8t2 (s = ut + 0.5at2) = 4.9t2 distance covered by rocketeer while he catches the student at t s d = Vo*(t-5) + 0.5*9.8(t-5)2 now 4.9t2 = Vo(t-5) + 4.9(t-5)2 ....eqn1 velocity of student at point of catching = gt a) let the minimum required height be h m above ground then let the upward acceleration be 5g velocity on landing v =0 v2 - u2 = 2as => 0 - g2t2 = 2*-5g*h => h = gt2/10 h+d = 553m => 4.9t2/10 + 4.9t2 = 553 =>t2 = 102.6 =>t = 10.13 s = 10s approx b) from eqn. 1 we get 4.9t2 = Vo(t-5) + 4.9(t-5)2 =>4.9*10^2 = Vo*5 + 4.9*5^2 => 5Vo = 4.9*75 => Vo = 73.5m/s
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