Two skydivers will step out of opposite sides of a stationary hot air balloon 45

Question

Two skydivers will step out of opposite sides of a stationary hot air balloon 457m above the ground. Assume that a free falling skydiver will fall with a constant acceleration of 9.8 m/s^2 before the parachute opens. The first skydiver falls with a constant speed of 3 m/s after opening her parachute. The second skydiver falls with a constant speed of 3.2 m/s after opening his parachute. If the first skydiver waits 4.0 seconds after stepping out of the balloon before opening her parachute, how long must the second skydiver wait after leaving the balloon before opening his parachute? From the time the first skydiver jumps, how long does the entire stunt take?

Explanation / Answer

S1=gt^2/2=80 m S2=457-80=377 m t2=S2/V=377/3=125.6 s t3=t2+t1=125.6+4=129.6s (Duration of stunt for 1 diver) S4=V*t4=3.2*125.6=401.92m S3=457-401.92=55.08m gt^2/2=55.08 t^2=11 t=sqrt(11)=3.3 s approx (the second skydiver wait after leaving the balloon before opening his parachute)

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