Asked by Natasha M.
A parachutist bails out and freely falls 59.7 m. Then the parachute opens and thereafter she decelerates at 1.60 m/s2. She reaches the ground with a speed of 3.41 m/s. What is her average speed (in m/s) for the fall?
Answers
Answered by
Henry
d1 = Vo*t1 + 0.5g*t1^2 = 59.7 m.
0 + 4.9t1^2 = 59.7.
t1^2 = 12.18.
t1 = 3.49 s.
V1 = Vo + g*t1 = 0 + 9.8*3.49 = 34.2 m/s.
V2^2 = V1^2 + 2a*d2 = 3.41^2 = 11.63.
34.2^2 - 3.2d2 = 11.63.
-3.2d2 = 11.63-1169.64 = -1158.
d2 = 362 m.
d2 = V1*t2 + 0.5a*t2^2 = 362.
34.2*t2 - 0.80*t2^2 = 362.
-0.8t2^2 + 34.2t2 - 362 = 0.
Use Quadratic Formula.
t2 = 19.28 s.
D = V*t.
V = D/T = (59.7+362)/(3.49+19.28) = 18.5
m/s.
0 + 4.9t1^2 = 59.7.
t1^2 = 12.18.
t1 = 3.49 s.
V1 = Vo + g*t1 = 0 + 9.8*3.49 = 34.2 m/s.
V2^2 = V1^2 + 2a*d2 = 3.41^2 = 11.63.
34.2^2 - 3.2d2 = 11.63.
-3.2d2 = 11.63-1169.64 = -1158.
d2 = 362 m.
d2 = V1*t2 + 0.5a*t2^2 = 362.
34.2*t2 - 0.80*t2^2 = 362.
-0.8t2^2 + 34.2t2 - 362 = 0.
Use Quadratic Formula.
t2 = 19.28 s.
D = V*t.
V = D/T = (59.7+362)/(3.49+19.28) = 18.5
m/s.
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