Question

A fighter jet takes off from the horizontal deck of an aircraft carrier. Starting from rest, it achieves a speed of 73.6 m/s as it rolls 85.6 m on the deck to the end of the runway, at which point the jet leaves the deck and its acceleration changes to 4.17 m/s² at 50.3 degrees above horizontal.

A- What is the magnitude of the acceleration, in m/s², of the jet while it is rolling on the deck?

B- What is the speed of the jet, in m/s, 4.38 s after it leaves the deck of the aircraft carrier?

c- What is the altitude (measured in meters, relative to the deck) of the jet 4.38 s after it leaves the deck of the aircraft carrier?

d- How far, horizontally in meters, is the jet from the end of the runway 4.38 s after it leaves the deck of the aircraft carrier?

Answer 1

(a) Use kinematic equation v^2 = u^2 + 2as   

u = 0 s = 85.6m v = 73.6m/s

putting it back we get

a = v^2/2s = 31.64 m/s^2

(b) Now the horizontal speed is 73.6 m/s

acceleration = 4.17m/s^2 at 50.3degrees

horizontal acceleration ax = 4.17cos(50.3) = 2.66 m/s^2

vertical acceleration ay = 4.17sin(50.3) = 3.21 m/s^2

Using kinematic equation v = u + at at t = 4.38 s

vertical speed vy = uy + ay*t = 0+3.21*4.38 = 14.05 m/s

horizontal speed vx = ux + ax*t = 73.6 + 2.66*4.38 = 85.25 m/s

total speed = sqrt(vx^2 + vy^2) = 86.4 m/s

(c) Using kinematic equation sy = uy*t + 1/2*ay*t^2

uy = 0

sy =0 + 1/2*ay*t^2 = 0.5*3.21*4.38^2 = 30.79 m

(d)

Using kinematic equation sx = ux*t + 1/2*ax*t^2

ux = 73.6

sy = 73.6*4.38 + 1/2*ax*t^2 = 347.88 m

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