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Motion of Model Rockets

by Andrew
(New York)










































A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 2.0 m/s² until its engines stop at an altitude of 150 m.

    (a) Describe the motion of the rocket after its engines stop.

    (b) Calculate the maximum height reached by the rocket.

    (c) How long is the rocket in the air?

Comments for Motion of Model Rockets

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Sep 15, 2012
Kinematics - Model Rockets
by: Staff


Answer:


Part I
 Graph of Model Rocket Altitude Vs. Time




(a) Describe the motion of the rocket after its engines stop.

          The rocket continues upward after its engines stop. However, its upward velocity decreases over time until all upward motion stops. At that point the rocket begins to fall back to the earth.


(b) Calculate the maximum height reached by the rocket.

          h = 158.1632653061226 m



(c) How long is the rocket in the air?

          Time in Air = 11.7606933414363 seconds



Definitions:


          h = maximum altitude in meters

          h₁ = altitude where engines stop

          h₂ = additional altitude achieved after engine stops

          t₁ = time in seconds until the engines stop

          t₂ = time in seconds that the rocket continues to travel upward after the engines stop

          t₃ = time it takes for the rocket to return to ground level from its maximum altitude

          vinitial = initial velocity

          vengine cut off = velocity at the time the engine is cut off

          afrom engines = acceleration from engines

          ggravitational pull = gravitational pull





Calculation of time to engine cut off:


      Constant linear acceleration

          afrom engines = 2.0 m/s² (upward)

          ggravitational pull = 9.8 m/s² (downward)

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Sep 15, 2012
Kinematics - Model Rockets
by: Staff


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Part II

      Initial upward velocity

          vinitial = 50 m/s


      altitude when engine cut off

          h₁ = 150 m


      time until engine cut off

          t₁ = unknown


      equation for distance traveled from launch until engine cut off

          distance (h₁)=initial velocity (vinitial)*time (t₁) + 1/2 net acceleration (a – g)*time² (t²₁)

          h₁ = vinitial * t₁ + ½ * (a – g) * t²₁


          h₁ = 50 * t₁ + ½ * (2 – 9.8) * t²₁


          150 = 50 * t₁ + ½ * (2 – 9.8) * t²₁

          150 = 50 * t₁ + ½ * (– 7.8) * t²₁

          150 = 50 * t₁ + (– 3.9) * t²₁

          150 = 50 * t₁ - 3.9 * t²₁

          150 - 150 = 50 * t₁ - 3.9 * t²₁ - 150

          -3.9 * t²₁+ 50 * t₁ - 150 = 0

          -3.9 t²₁+ 50 t₁ - 150 = 0

      This is a quadratic equation. You can solve for t₁ using the quadratic formula.

          a t²₁ + b t₁ + c = 0

          t₁ = [-b ± √(b² - 4ac)]/(2a)

          t₁ = [-50 ± √((-50)² - 4*(-3.9)*(-150))]/[2*(-3.9)]

          t₁ ∈ (4.788575558888, 8.0319372616248 )

          t₁ cannot be two different values. The lower value is the correct answer since the rocket motors stop when the rocket reaches 150 the first time. After t₁ = 4.788575558888 seconds, the acceleration changes to -9.8 m/s² (rather than -7.8 m/s².


          (However, note that if the acceleration afrom engines continued throughout the entire flight, the second value (8.0319372616248) would also be correct. Even with afrom engines the rocket would still fall back to earth. It would pass through 150 m altitude a second time as it descended.)

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Sep 15, 2012
Kinematics - Model Rockets
by: Staff


-------------------------------------

Part III


          t₁ = 4.788575558888 seconds



Calculation of velocity at engine cut off:

          vengine cut off = 50 + (2 – 9.8) * t₁

          vengine cut off = 50 + (2 – 9.8) * 4.788575558888

          vengine cut off = 12.6491106406736 m/s

          vengine cut off = 12.6491106406736 m/s



Calculation of additional time t₂ needed to achieve maximum altitude (when velocity is reduced to zero):

          velocity = 12.6491106406736 - 9.8 t₂

          0 = 12.6491106406736 - 9.8 t₂

          9.8 t₂ = 12.6491106406736

          9.8 t₂ / 9.8 = 12.6491106406736 / 9.8

          t₂ * (9.8 / 9.8) = 12.6491106406736 / 9.8

          t₂ * (1) = 12.6491106406736 / 9.8

          t₂ = 12.6491106406736 / 9.8

          t₂ = 1.2907255755789 seconds

          t₂ = 1.2907255755789 seconds



Calculation of maximum altitude:

          h₂ = 12.6491106406736 * t₂ + ½ * ( – 9.8) * t²₂

          h₂ = 12.6491106406736 * (1.2907255755789) + ½ * ( – 9.8) * (1.2907255755789)²

          h₂ = 12.6491106406736 * (1.2907255755789) + ½ * ( – 9.8) * (1.2907255755789)²

          h₂ = 8.1632653061226



          h = Maximum Altitude = h₁ + h₂

          h = 150 + 8.1632653061226

          h = 158.1632653061226 m

          h = 158.1632653061226 m


-------------------------------------

Sep 15, 2012
Kinematics - Model Rockets
by: Staff


-------------------------------------

Part IV



Calculation of time t₃ it takes for the rocket to return to ground level from its maximum altitude:


          h = ½ * (– g) * t²₃


          h = ½ * (– g) * t²₃

          h = -½ * g * t²₃

          158.1632653061226 = -½ * 9.8 * t²₃

          2*158.1632653061226 = 2*-½ * 9.8 * t²₃

          2*158.1632653061226 = -2*½ * 9.8 * t²₃

          2*158.1632653061226 = -(2/2) * 9.8 * t²₃

          2*158.1632653061226 = -(1) * 9.8 * t²₃

          2*158.1632653061226 = 9.8 * t²₃

          316.3265306122452 = 9.8 * t²₃

          316.3265306122452 / 9.8 = 9.8 * t²₃ / 9.8

          316.3265306122452 / 9.8 = t²₃ * (9.8 / 9.8)

          316.3265306122452 / 9.8 = t²₃ * (1)

          316.3265306122452 / 9.8 = t²₃

          32.2782174094128 = t²₃

          t²₃ = 32.2782174094128

          √(t²₃) = √(32.2782174094128)

          t₃ = 5.6813922069694


          t₃ = 5.6813922069694 seconds



Calculation of time the rocket is in the air:

          Time = t₁ + t₂ + t₃

          Time = 4.788575558888 + 1.2907255755789 + 5.6813922069694

          Time = 11.7606933414363


          Time in Air = 11.7606933414363 seconds





Thanks for writing.

Staff
www.solving-math-problems.com


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