Exercise 1 Rectilinear Motion

Exercise 1: Rectilinear Motion
Background
The average accelerations of a US Navy fighter jet takeoff to an US Air Force fighter jet takeoff are very different. The aircraft conditions are the same except the Air Force fighter is on a 12,000-ft dry concrete runway and the Navy fighter is on the catapult of an aircraft carrier. (Hint: You will have to use two different methods because you are not given the catapult force, but you have the catapult distance. Assume aircraft is airborne after the catapult stroke.)
Givens:
Gross Weight (W) = 55,000 lb
Average Drag (D) = 9,000 lb
Average Friction (f) = 1,500 lb
Average Thrust (T) = 36,000 lb
Lift Off Speed ( VLO) 180 KTAS
Catapult Distance = 350 ft
1. Compute the average acceleration of the Air Force Fighter during the takeoff roll (ft/s 2).
F = ma
36000 – 9000 – 1500 = (55000/32.17) * a a = 14.92 ft/s2
2. What would be the length of the takeoff roll for the Air Force Fighter (ft)?
V2 = U2 + 2aS
(180*1.6878)2 = 0 + 2*14.92*S
S = 3093.06 ft
3. How long (time) would it take for the Air Force Fighter to liftoff (sec)?
V = U + at
(180*1.6878) = 0 + (14.92)(t) t = 20.4s
4. Compute the average acceleration of the Navy Fighter during catapult launch (ft/s2).
V2 = U2 + 2aS
(180*1.6878)2 = 0 + 2(a)(350) a = 131.85 ft/s2
5. How long (time in seconds) does it take for the Navy fighter to takeoff with a catapult launch?
V = U + at
(180*1.6878) = 0 + 131.85(t) t = 2.3s
6. Find potential, kinetic, and total energy of this aircraft after takeoff at the following conditions:
Weight = 50,000 lb
Altitude = 10,000 ft
Airspeed = 300 KTAS
PE = mgh
PE = (50000)(10000)
PE = 5*108
KE = ½mv2
KE = ½(50000/32.17)(300*1.6878)2
KE = 1.99*108
TE = 6.99*108
7. Given conditions of #6, find new altitude if the aircraft raises the nose to capture 250 KTAS but keeps the thrust at the same setting and, for simplicity, assume the drag stays the same. Think of this as a Constant Energy Problem.
TE = KE + PE