Question

If we dropped a car weighing 4451 lbs. from a C-130 aircraft at 5,280 ft, how much horsepower would it take to drive past it before it hits the ground if you’re 1 mile away?

Pro Tips

Air density @ sea level, 59 degrees, no wind = p = .002377 slugs/ft^3

Coefficient of drag (flat plate, NASA) = C(d) = 1.28

Weight = W = 4451 lbs

Gravitation constant = g = 32.2 ft/sec^2

Area = A = 197.5"" long x 78.2"" wide x (1 ft^2/ 14 in^2)

Vehicle falls flat, wheels 1st, straight down, at constant acceleration with no aerodynamic drag until terminal velocity

Horsepower needed to accelerate is AVERAGE - not peak

100% driveline efficiency

Answer 1

Gravitational force on the car = mg.

air drag force = 1/2 v²C_d A.

When the car reached terminal velocity both force on the car will be equal

mg = 1/2 v²C_d A.

terminal vel . v_t² = (2mg/(C_d A.) )

= 2*4451/ (0.002377*1.28*107.25) ) = 27280

v_t = 165.17 ft/s

h is fall in height , fall at const. g=32 ft until v _t is reached , initial vel =0 at h= 4451 ft

v² = 2gh

h = 27280/32.2*2 = 423.6

the car has fallen through 852.5 ft and reached terminal velocity v_t = 165.17 ft. The it falls

5280 - 423.6 = 4856.5 ft with const. vel of v_t

time of fall t = 4656.5 /165.17 = 29.4 s

You have drive 1 mile(5280ft) 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*29 = 4625 HP