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Drag Force Question

  • Thread starter bibisan
  • Start date
  • #1
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Can anyone Please help me on this. I am totally stuck at this question :surprised .
thanks.....

...A sky diver of mass 80.0 (including parachute) jumps off a plane and begins her descent.
Throughout this problem use 9.80 for the magnitude of the acceleration due to gravity.

Q1.
At the beginning of her fall, does the sky diver have an acceleration?
----> is it : Yes and her acceleration is directed downward.

Q2.
When the sky diver descends to a certain height from the ground, she deploys her parachute to ensure a safe landing. (Usually the parachute is deployed when the sky diver reaches an altitude of about 900m--3000ft.) Immediately after deploying the parachute, does the skydiver have a nonzero acceleration?
------> is it:Yes and her acceleration is directed upward.

Q3.
When the parachute is fully open, the effective drag coefficient (k) of the sky diver plus parachute increases 0.25 to 60.0kg/m . What is the drag force, FD acting on the sky diver immediately after she has opened the parachute? (given F=kv^2)

Q4.
What is the terminal speed of the sky diver when the parachute is opened?
 

Answers and Replies

  • #2
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(Part A) Yes, and her acceleration is directed DOWNWARDS.

(Part B) This will be equal to her weight = 80(9.8) = 784N.

(Part C) v^2 = 784/0.25 = 56m/s.

(Part D) Yes, and her acceleration is directed UPWARDS.

(Part E) F = 60 ( 56 ) = 3360 N.

(Part F) Applying the equation F=Kv^2,

the key I used here is that the velocity will reduce to a value such that the Force F will equal the weight.

Thus, v^2 = 784/60 = 3.614m/s.
 

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