Asked by Megan
A bobcat weighing 74.5 N jumps out of a tree. (Assume that "up" is the positive direction.)
(b) What is the drag force on the bobcat when it falls at 70% of its terminal velocity?
(d) What is the acceleration when it falls at 70% of its terminal velocity?
(b) What is the drag force on the bobcat when it falls at 70% of its terminal velocity?
(d) What is the acceleration when it falls at 70% of its terminal velocity?
Answers
Answered by
drwls
(b) Drag force is proportional to V^2. At the full terminal velocity, the drag force equals the weight. At 70% of that velocity, the drag force is 0.70^2 = 0.49 times the weight. That is 36.5 N.
(d) Net force = 0.51 M g = M a
M cancels out.
Solve for a
(d) Net force = 0.51 M g = M a
M cancels out.
Solve for a
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