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Flying Rocket

  1. Apr 3, 2005 #1
    Q. In an experiment in physics class, a model solid-fueled rocket is fired vertically with an acceleration of 3 m/sec2 for 5 seconds. After that time, it's fuel is exhausted and it continues upwards as a free fall particle. (Take upwards to be the positive direction.)

    a) What is the maximum altitude the rocket reaches?

    Ans. x-x0=vo(t)+(1/2)a(t)^2
    where x0=0
    a=3 m/sec^2

    I get x= 37.5 meters

    v=0, a=-9.81m/s/s t=5 sec

    Using v = v0 + at, we get

    v0 = 49.05 m/s^2

    While the rocket engine is firing a= 3. v0= 0 so v= 0+ 3(5)= 15 and you can use THAT as v0 after the engine shuts down.

    AFTER the rocket engine has shut off, the only acceleration is that of gravity: -9.81 so v= 15- 9.81t (t is now measured from engine cutoff, not initial launch). From that we get h= 15t- 4.9t2+ 122.5 (since the height was 37.5 m at engine cutoff, t=0).

    The maximum height occurs when v= 0: 15- 9.81t= 0 gives t= 15/9.81= 1.52 seconds. That height will be h= 15(1.52)- 4.9(1.52)^2+ 37.5= 48.97 m.

    b) What is the rocket's acceleration 6.5 seconds after the launch?

    Here is where I am stuck. Any ideas....

    c) What is the total time of flight?

    Some ideas:

    Finally, the flight will be over when the rocket hits the ground:
    h=0. Solve 15t- 4.9t2+ 37.5= 0 to find the time AFTER CUTOFF that that occurs. You will need to add the inital 5 seconds to find the length of the entire flight.
    I tried to do this but my answer is wrong. ( The computer does not take it )

    d) What is the rocket's total displacment during the entire flight?

    The answer is zero, but can anybody explain why?
  2. jcsd
  3. Apr 3, 2005 #2
    Displacement is not the total distance moved, but a vector of the distance from the starting point to the endpoint. Since it ended up in the same place it started (on the ground), the displacement is 0.

    Look up the exact definition of displacement in a physics textbook if that doesn't quite make sense.
  4. Apr 3, 2005 #3
    Can anybody help me on part b and c.... Plz
  5. Apr 3, 2005 #4
    b) It is no longer accelerating upwards, what forces are acting on it as a free fall particle? Whats the magnitude of this force?

    c) Find the time it takes to get to the top. The time for it to fall will just be a freefall problem with the kinetic equation for freefall.

    [tex] t = \sqrt{\frac{2h}{g}} [/tex]
  6. Apr 3, 2005 #5
    b) Acceleration due to gravity is acting only

    I tried doing, v = v0 + at

    v0 =0

    a = v / t
    v = 49.05 m/s
    t = 6.5 s

    a= 7.54. which is wrong????

    c) I used t = sqrt (2h/g) 3.15 seconds, + the Orignal t = 5 seconds = 8.15 seconds which is wrong.

    Don't know what is wrong here..........
  7. Apr 4, 2005 #6
    If the only force acting is gravity, then dividing by the mass will give you the acceleration. Gravity is the only source of acceleration, and gravity is 9.8m/s^2.

    It was accelerating for 5 seconds, that doesn't mean it took 5 seconds to get to the top of its flight. Try another approach.
  8. Apr 4, 2005 #7
    No calculations are needed for b). The acceleration due to gravity is a constant for a given gravitational field.

    In part c) 5 seconds is the time till the feul runs out, not the time to the maximum height. Since the initial upward acceleration is consant the velocity when the fuel runs out is 3*5 = 15 m/s use this as vo in v = vo - gt, and solve for t at v = 0. This will give you the time from fuel exhaustion to maximum height.
  9. Apr 4, 2005 #8
    Still didn't get this can anyone help me with this

    wht exactly to substitute......

    Plz. help!!
  10. Apr 4, 2005 #9
    Part c)

    Upward acceleration of 3m/s^2 for 5 secs. To find the velocity acquired from this acceleration, use the equation

    [tex] v = a t = (3) (5) = 15m/s [/tex] This is the velocity when the fuel runs out. From here the only force acting is gravity, which is -9.8m/s^2
    Find the displacement from this acceleration:
    [tex] x = at^2/2 = (3)(5)^2/2 = 75/2 = 37.5m [/tex] This is how high it went when it ran out of fuel.

    Using the position equation to find how long it takes to fall back down:
    [tex] x = x_0 + v_0t + \frac{gt^2}{2} [/tex]

    Total displacement is -37.5m
    Initial velocity = +15m/s
    Acceleration = -9.8m/s^2

    [tex]0 = -4.9t^2+15t+37.5[/tex] If you solve this, you'll find the total time of flight from the point when the fuel runs out. Add 5 to this to find the total time. The solutions for t to this equation are

    [tex] t = \frac{2v-\sqrt{v^2+2gh}}{2g} [/tex] and [tex] t = \frac{2v+\sqrt{v^2+2gh}}{2g} [/tex]
  11. Apr 4, 2005 #10
    Notice that your answer for a is incorrect. It travels 37.5m in 5 seconds, but at 5 seconds the acceleration stops. It still has upward momentum, and is still going upwards.

    Energy analysis might be easier to solve this problem, but I have class soon :/
  12. Apr 4, 2005 #11
    Now, can somebody help me on part b,

    Some explanation please.
  13. Apr 4, 2005 #12
    Part b is the easiest part. It doesnt matter whats happened at any time before or what will happen after. At any time, the force of a body in earth's gravitational field will be 9.8*m. Whether it was accelerating before or after is irrelevant.

    If nothing ihappening, then assume the normal condition that gravity is pulling down.
  14. Apr 4, 2005 #13
    But how to find m.

    F = ma

    m = F/a

    m = 9.81 / 9.81
    So, a = F/m

    = 9.81 / 1 = 9.81, Is this right , then
  15. Apr 4, 2005 #14
    Yes that's right. The only force your considering acting on the rocket is gravity, so

    rockets acceleration = g
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