Given g=9.8 m/s^2. A sky diver of mass 65 kg jumps from a slowly moving aircraft

Question

Given g=9.8 m/s^2. A sky diver of mass 65 kg jumps from a slowly moving aircraft and eventually reaches her terminal velocity 68 m/s. What was (the magnitude of) her acceleration at the moment her velocity was only V1 = 45.4 m/s (directed vertically down)? Answer in units of m/s^2
Now consider the same sky diver jumping out of a faster airplane. At some point before she reaches her terminal velocity, she has velocity V2 of magnitude 83.7 m/s in the direction 25 degrees below horizontal. The magnitude of the sky diver's acceleration at that point in time is
(9.8-(1/2 x (2x9.8) /68^2) x 45.4^2 = 5.43 m/s^2
Now consider the same sky diver jumping out of a faster airplane. At some point before she reaches her terminal velocity, she has velocity V2 of magnitude 83.7 m/s in the direction 25 degrees below horizontal. What is the magnitude of the sky divers acceleration at that point in time? Answer in units of m/s^2
(I have the equation down to
mg-(1/2 (2mg/Vt^2) V^2) = ma
Need help plugging it in....

Explanation / Answer

so a=9.8 so t=v/a=68/9.8 now acc at v=45.4=(9.8-9.8*45.4/68)=3.26

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