Question

The stories of a tall building are uniformly 10.0 feet in height. A ball A is dropped from the rooftop position shown. Determine the times required for it to pass the 10.0 feet of the first, tenth, and one-hundredth stories (counted from the top). Include the effects of aerodynamic drag. The drag force causes an acceleration component in ft/sec2 of 0.005v2 in the direction opposite the velocity vector, where v is in ft/sec 10.0 A

Answer 1

Ciren building. loft per storey 0.005v2 retardation. fbd of Ball dstepped from A 9 acceleration So initial uzo atto. So Net accelaration a=g-0.00512 a= 32.2 -0.00512 dx = 32-2-0-00 sve {9=32-24/sec) dv Integraty both side Integooting boty side 32 a osuz afat 2-0.00 SUZ - 200 dv. J a²-x² 6440 - V2 + 200 per = + 1.2461) en vento ty -t berte EV = 6-8025)+ 2v RJ6440 0.8025 t - 68025 tt v = (80.25).po 50.8025t-11 T 0.8025 t t .

bost - 0.8025 t -, 1 80.25 0-8025 + + for first floor. integrating Y=lott. . time=? so 0-8025t-1 at 0.8025€ +1 dy = Jo.25 + 1.246 Vot 2-37 38 t+1.246 10 = 80.25 10 = 80.25j te tillat 10 = 80-254**- e n te 10 = 80.24 [+ - 2.432_luft+p240)* 10 = 80-24 [t; -2.492 ln(t+1-240) +2.492 lut I Solving too hit and trial.. = 1.5762.see for 18th floor. Limit ya oto 100ft. 100= 80.24[to-2.492 encerts +1246) +2.492 en tro) Again solving fortlight and trial. O Eto = 2.318193602] e for 100th floor. Uimit yť o to loooft. 1000 = 80.24[too -2.4922n(tie ti-244 +2.492 24 theo . Solving for two by hit and trial Hros = 12.695 ste