A parachutist in vertical free fall opens the parachute when the speed is v0. Th

Question

A parachutist in vertical free fall opens the parachute when the speed is v0. The total mass of the parachute plus person is m. Assume that the quadratic model of air resistance is appropriate during the motion, and take the effective diameter of the parachute and parachutist to be D (so that the air resistance force has magnitude c2D2v 2 , where v is the velocity of the parachutist and c2 is a constant). Assume that the parachutist continues to fall vertically, and take the magnitude of acceleration due to gravity to be g.

(a) Define an appropriate coordinate axis, draw a force diagram and write down the vector form of all forces.

(b) Apply Newtonian mechanics and present the equation of motion for the parachutist.

(c) Solve the terminal speed vT of the parachutist in terms of m, g, D and c2. Hence show that the equation of motion may be expressed as v dy/dx = -g/vT^2 (v^2 - vT^2)

(d) Initially the parachutist has speed v0 > vT . Find, in terms of v, vT , v0 and g, the distance fallen after the parachute opens.

Explanation / Answer

orient the y axis downwards.

Fair = - c2v2D2

Fg = mg

ma = mg - cv2D2

m dv/dt = mg - c2v2D2

when terminal velocity is reached, dv/dt = 0 and therefore

VT = (mg)1/2/cD

Substituting back for cD

m v dy/dx = mg - v2[mg/vt2]

vdy/dx = g(Vt2 - v2)/Vt2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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