An 82.0 kg skydiver jumps out of a balloon at an altitude of 1 280 m and opens t

Question

An 82.0 kg skydiver jumps out of a balloon at an altitude of 1 280 m and opens the parachute at an altitude of 190 m. (a) Assuming that the total retarding force on the diver is constant at 70.0 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 sky diver will be injured? Explain. c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s? __________m/s (d) How realistic is the assumption that the total retarding force is constant? Explain your answer.

Explanation / Answer

(a) The net force on the skydiver before the parachute opens is: F(net) = F(retarding force) - F(gravity) = 70 N - (82.0 kg)(9.8 m/s^2) = -733.6 N. So, the acceleration of the skydiver is: a = F(net)/m = (-733.6 N)/(82.0 kg) = -8.95 m/s^2. Using (v_f)^2 = (v_i)^2 + 2ad, the speed of the diver when he opens his parachute is: v_f = ?[(v_i)^2 + 2ad] = ?[(0.0 m/s)^2 + 2(8.95 m/s^2)(1280-190) m)] = 139.68 m/s. At this point, you can find the speed of the skydiver after he opens his parachute

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.