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

 Sep 15, 2012 Kinematics - Model Rockets by: Staff Answer: Part I (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) -------------------------------------

 Sep 15, 2012 Kinematics - Model Rockets by: Staff ------------------------------------- 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.) -------------------------------------

 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