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A sky diver jumps from a helicopter and is falling straight down at 30 m/s when his parachute opens. From then on, his downward acceleration is approximately

a = g - cv2 , where g = 9.81 m/s2 and c is a constant. After an initial "transient" period, he descends at a nearly constant velocity of 5 m/s.


a. What is the value of c, and what are its SI units?


b. What maximum deceleration is the sky diver subjected to?


c. What is his downward velocity whe he has fallen 2m from the point at which his parachute opens?

Respuesta :

akande212

Answer:

(a) c = 0.392/m

(b) a = 9.80m/s²

Explanation:

Given

a = g - cv2 = 9.8 – cv²

When velocity is constant acceleration a = 0. So, for a constant velocity of 5m/s

0 = 9.8 – c(5.0)²

0 = 9.8 – c×25

0 = 9.8 – 25c

25c = 9.8

c = 9.8/25 = 0.392/m SI unit of c is (/m)

(b)

Maximum deceleration occurs when da/dv = 0

So differentiating a with respect to v we have that

da/dv = 0 = 2cv

2×0.392v = 0

0.784v = 0

v = 0

So the maximum deceleration occurs when v = 0m/s this corresponds to the time when the sky diver initially jumps down from the helicopter as his in free fall.

So amax = 9.80 -0.392v² =9.80 - 0.392(0)²

amax = 9.80m/s² = g just confirms the reasoning above.

(c)

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