An 80.0-kg skydiver jumps out of a balloon at an altitude of 1,010 m and opens t

Question

An 80.0-kg skydiver jumps out of a balloon at an altitude of 1,010 m and opens the parachute at an altitude of 180.0 m.

(a) Assuming that the total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3,600 N with the parachute open, what is the speed of the diver when he lands on the ground?
___________ m/s

(b) Do you think the skydiver will get hurt? Explain.

(c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.80 m/s?
__________m

d) How realistic is the assumption that the total retarding force is constant? Explain.

Explanation / Answer

weight W = mg = 784.8 N

While the parachute is closed, the net force is
Fnet = 784.8N - 50N = 698.8N
so the acceleration a = Fnet / m = 9.985 m/s²
v² = u² + 2as = 0 + 2 * 9.985m/s² * (1010m - 180m) = 16575.1 m²/s²
v = 128 m/s

While the parachute is open, the net force is
Fnet = 748.8N - 3600N = -2851.2 N
so the acceleration a = -2851N / 80kg = -35.64 m/s²
v² = (128m/s)² - 2 * 35.64m/s² * 180m = 3553.6 m²/s²
v = 59.60 m/s

(b) Hurt? Most definitely. That's over 59.60 mph.

(c) Work it backwards:
(5.8m/s)² = u² - 2 * 35.64m/s² * s
But u² = 2 * 9.985m/s² * (1010m - s) = 20169.7 - 19.97*s. Plug that in and simplify:
33.64 = 20169.7 - 19.97s - 71.28s
104.92s = 20149.73
s = 192.04 m

(d) Not very realistic -- the retarding force is proportional to the square of the velocity.

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