Question

Drop a 4451 lb flat plate (area = 15444.5 sq inches) which stays flat while falling from 1 mile high. How much HP needed for a car to drive past impact point starting at zero and 1 mile from impact point? Known factors: Coefficent of drag C(d)= 1.28 , Air Density (no wind) p = .002377slugs per ft cubed. Gravity constant =g=32.2ft/sec squared, Horsepower needed to accelerate is AVERAGE - not peak

Answer 1

• Gravitational force on the plate = mg.

  air drag force = 1/2 $\rho$ v²C_d A.

• area of the car A = 15444.5 /144 = 107.25 sq/ft

  C_d = 1.28
• $\rho$ = 0.002377 slug/cft = 0.002377*32.2 lb/cft
• drag force F_d = 1/2 $\rho$ v²C_d A. = 5.25v²

The plate is subjected to gravity and sir drag

its motion

ma = mg - F_d

dv/dt = g- 5.25v²

dv/( 1.87-v²) = 5.25 dt

integrating both sides we get :

use identity

u”? – 70 – zB 0-2 "T rp

V – 1.37 V +1.37 = -14.190

v(t) = 1371 +e-14.194 1-e14.19

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

  mg = 1/2 $\rho$ v²C_d A.

  terminal vel . v_t² = (2mg/($\rho$C_d A.) )

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

  v_t = 165.17 ft/s

• put v(t) = 165.17 in the above expression we get t = 0.001s

  it will reach terminal velocity at 0.001 s ,can be ignored

  time of fall t = 5280 /165.17 = 31.97 s

  You have to drive 1 mile (5280 ft) in less than 31.97 s to drive past before it falls to ground , starting with 0 vel

• take it 31 s on safer side

  speed required = 5280/31 = 170.3 ft /s - average velocity

  we take mass of the car mass as 4451 lbs.

  KE of the car = 1/2 mv² = 1/2 4451*(170.3)² = 64,561,269 ft.lb

  you have to drive in 31 s

  1 HP = 550 ft.lb/s

  HP required to drive =64,561,269/550*31 = 3787 HP