In: Physics
Gravitational force on the car = mg.
air drag force = 1/2 v2Cd A.
When the car reached terminal velocity both force on the car will be equal
mg = 1/2 v2Cd A.
A = 197.5*78.2 /144 = 107.25 sq.ft
terminal vel . vt2 = (2mg/(Cd A.) )
= 2*4451/ (0.002377*1.28*107.25) ) = 27280
vt = 165.17 ft/s
h is fall in height , fall at const. g=32.2 ft until v t is reached , initial vel =0 at h= 4451 ft
v2 = 2gh
h = 27280/32*2 = 423.6 ft
the car has fallen through 852.5 ft and reached terminal velocity vt = 165.17 ft. The it falls
5280 - 423.6 = 4856.4 ft with const. vel of vt
time of fall t = 4856.4 /165.17 = 29.4 s
You have to drive 1 mile (5280 ft) in less than 29.4 s to drive past before it falls to ground
take it 29 s on safer side
speed required = 5280/29 = 182.1 ft /s
we take mass of the car mass as 4451 lbs.
KE of the car = 1/2 mv2 = 1/2 4451*(182.1)2 =73,773,340 ft.lb
you have to drive in 29 s
1 HP = 550 ft.lb/s
HP required to drive =73,773,340/550*26 = 4626 HP