Question

What is the magnitude of the sky diver's acceleration at the instant the parachute opens? Express your answer with the appropriate units. Constants A sky diver and his parachute weigh 200 lb. He is falling vertically at 80 ft/s when his parachute opens. With the parachute open, the magnitude of the drag force (in pounds) is 0.5u2 ft Figure a 16.2 1 of 1 ST Submit X Incorrect; Try Again; 4 attempts remaining Part B What is the magnitude of his velocity when he has descended 22 ft from the point where his parachute opens? Figure 1) Express your answer with the appropriate units.

Answer 1

At instant the parachute opens F_drag = 0.5 V^2 = 0.5 times 80^2 = 3200 F_net = 3200 - 200 = 3000 lb acceleration = F_net/m = 3000/(200/32.2) = 483 ft/s^2 \r\n F_net = W - F_d F_net = 200 - 0.5 V^2 ma = 200 - 0.5 V^2 (200/32.2) V. dv/ds = 200 - 0.5 V^2 6.211 V dv/ds = (200 - 0.5 V^2) 6.211 integral V dv/200 - 0.5 v^2 = integral ds put 200 - 0.5 V^2 = t - v dv = dt or integral v dv/200 - 0.5 v^2 = -integral dt/t = -ln t -6.211 ln (200 - 0.5 v^2) = S + C \r\n F_net = W - F_d F_net = 200 - 0.5 v^2 ma = 200 - 0.5 v^2 (200/32.2) V dv/ds = (200 - 0.5 v^2) 6.211 integral v dv/200 - 0.5 v^2 = integral ds put 200 -0.5 v^2 = t - v dv = dt or integral v dv/200 - 0.5 v^2 = -integral dt/t = -lnt -6.211 ln (200 - 0.5 v^2) = s + c \r\n At s = 0, v = 80 ft/s -6.211 ln (200 - 0.5 times 80^2) = 0 + C