Question

An airplane is flying at a constant horizontal velocity of 184.5 m/s and drops a rocket. The rocket falls for 10.24 s without air resistance and under the influence of gravity only, and then fires its thrusters. From the moment the thrusters are one, the rocket has a net acceleration of 5.481 direscted 30.21 degrees above the horizontal. How far apart are the plane and the rocket when the rocket passes back through the cruising altitude of the plane?

Answer 1

for the rocket during falling

distance fallen y = (1/2)*g*t^2 = (1/2)*9.8*10.24^2 = 513.8 m

velocity of rocket vy = voy + ay*t

vy = 0 - 9.8*10.24 = -100.352 m/s

along horizontal

distance travelled x = vo*t = 184.5*10.24 = 1889.28 m

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after the rocket is fired

along vertical

displacement y = 513.8 m

initial velocity vy =-100.352 m/s

acceleration ay = 5.481*sin30.21 = 2.76 m/s^2

y = vy*t1 + (1/2)*ay*t1^^2

513.8 = -100.352*t1 + (1/2)*2.76*t1^2

t1 = 77.52 s

along horizontal

initial velocity vox = vo = 184.5 m/s

acceleration ax = a*cos30.21 = 4.75 m/s^2

distance travelled x1 = vox*t1 + (1/2)*ax*t1^2

x1 = 184.5*77.52 + (1/2)*4.75*77.52^2 = 28574.65 m

total horizontal distance travelled by rocket after reaching the cruising altitude of plane

s1 = 1889.28 + 28574.65 = 30463.93 m

for the airplane

distance travelled s2 = vo*(10.24+77.52) = 184.5*(10.24+77.52) = 16191.72 m

seperation between rocket and plane = s1 - s2 = 30463.93 - 16191.72 =14272.21 m