Skydiver #1 from the University Skydiving Club steps out of a planewhen it is 1

Question

Skydiver #1 from the University Skydiving Club steps out of a planewhen it is 1 mile above the ground and 10 seconds later skydiver #2steps out of the plane (at the same height). They both want to landon the ground at the same time. To make an estimate assume that askydiver falls with a constant assceleration of 32 ft/s2before the parchute opens and once the parachute has opened theskydiver falls with a constant velocity of 12ft/s. if skydiver#1opens his parachute 2 seconds after stepping out of the pane; howlong should skydriver #2 wait before opening her parachute?

Please show all steps for a lifesaver rating.

Explanation / Answer

Let the skydiver 2 open the chute after tseconds. So, for skydiver 1, time to reach ground be t1 h - ho = 0.5*a*t*t So, in 2 seconds with acceleration 32 ft/s2, d1 = 0.5*a*2*2 = 0.5*32*2*2 ft Now, after opening the parachute, d2 = v*(t1-2) = 12*(t1-2) So, d = d1+d2 = 64 + 12(t1-2) = 40 + 12*t1 ft Now, for skydiver 2,    d1 under acceleration = 0.5*32*t*t =16*t*t ft    d2 after opening chute = 12*(t2-t)    => d = d1+d2 = 16*t*t - 12*t + 12*t2    where t2 is the total time after jumping off theplane for skydiver 2. Also, t1-t2 = 10s So, 40 + 12*t1 = 16*t*t - 12*t + 12*(t1-10) => 40 = 16*t*t - 12*t - 120 => 4t*t - 3t - 40 = 0 => t = [3 ±(9 + 640)]/(2*4) s =3.56seconds

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