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Homework Help: How do I calculate the height this water rocket will rise?

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A 15.0 g plastic water rocket containing 40.0 g of water and compressed air is strung on a very low friction horizontal wire. When fired, compressed air forces the water from the rocket in 0.600 s at an average exit velocity of 4.05 m/s [E].

    Based on the information in the above narrative, if the rocket had been directed upwards, approximately how high would it have risen?

    The answer is supposed to be 6 m high, but I cannot figure out how to get it.

    2. Relevant equations
    V = d/t
    a = v/t
    a = Fnet/m
    g = Fg/m
    d = vi*t + 1/2at^2
    d = vf*t - 1/2at^2
    d = ((vf + vi)/2)t
    vf^2 = vi^2 + 2ad
    Ek = 1/2mv^2
    Ep = mgh
    W = delta E

    3. The attempt at a solution
    First I calculated the acceleration of the water with the formula a = v/t, and it gave me 6.75 m/s^2 [E].
    4.05 m/s [E] / 0.600 = 6.75 m/s^2 [E]

    Then I changed all the masses into kg.
    15.0 g = 0.015 kg
    40.0 g = 0.04 kg

    Then I calculated the net force of the water with the formula F = ma, and it gave me 0.27 N [E].
    0.04 kg * 6.75 m/s^2 [E] = 0.27 N [E]

    Newtons law states that for every action, there is an equal and opposite reaction, meaning that there would also a be a force of 0.27 N to the west. (The force that would make the rocket propel.)

    I then calculated the acceleration of the rocket using the formula a = Fnet/m, giving me 18 m/s^2 [W].
    0.27 [W] / 0.015 kg = 18 m/s^2 [W]

    If the rocket is directed upwards, it is now fighting against gravity, so I subtracted 9.81 m/s^2 [D] from my 18 m/s^2 [Up], giving me 8.19 m/s^2 [Up].
    18 m/s^2 [Up] - 9.81 m/s^2 [D] = 8.19 m/s^2 [Up]

    Using the formula d = Vi * t + 1/2at^2, I calculated the distance between the ground and the point when the rocket finished propelling all of its water out, giving me 1.47 m.
    0 m/s [Up] * 0.600 s + 1/2( 8.19 m/s^2 [Up] * 0.600 s^2 ) = 1.47 m

    Using the formula Vf = Vi + at, I calculated the velocity when the rocket finished propelling all of its water out, giving me 4.92 m/s [Up].
    0 m/s [Up] + 8.19 m/s^2 [Up] * 0.600 s = 4.92 m/s [Up]

    I then used the formula d = Vf^2 - ( Vi^2 / 2a ) to calculate the distance the rocket when up after propelling all of its water out.
    0 m/s [Up] ^2 - ( 4.92 m/s [Up]^2 / (2 * 9.81 m/s^2 [D] )) = 1.23 m

    I then added the two distances together, giving me 2.7 m. Apparently, this is incorrect.
    1.47 m + 1.23 m = 2.7 m.
  2. jcsd
  3. Feb 22, 2015 #2


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    Homework Helper

    Welcome to PF

    It may be simpler if you find the initial velocity (after the 0.6 seconds) from momentums instead of forces. What is the momentum of the system after the 0.6 seconds? What is the momentum of just the rocket?
  4. Feb 22, 2015 #3
    Thanks for the quick reply.

    By using the formula P = mv, I am able to find the momentum of the water leaving rocket.
    0.04 kg * 4.05 m/s [E] = 0.162 kg*m/s [E]

    I am assuming that momentum is similar to force and would give the the rocket a momentum of 0.162 kg*m/s [W].
    I could then find the velocity of the rocket with the formula v = P/m.
    0.162 kg*m/s [W] / 0.015 kg = 10.8 m/s [W].

    I used the formula d = Vf^2 - ( Vi^2 / 2a ) to calculate the distance after all the water was emptied out.
    0 m/s [Up] ^2 - ( 10.8 m/s [Up]^2 / (2 * 9.81 m/s^2 [D] )) = 5.94 m

    I added the two distances together, and again I did not get 6 m. At least it is closer than my last attempt.
    5.94 m + 1.47 m = 7.41 m

    The 5.94 m by itself is the correct answer, but shouldn't that be the distance it moved after the 0.6 seconds?
  5. Feb 22, 2015 #4


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    Science Advisor
    Homework Helper
    Gold Member

    A classic case of plugging in numbers just because they're the right units for an equation, without regard to what the equation actually requires the numbers to mean.
    Does the water accelerate from 0 to 4.05 m/s gradually over a period of 0.6 seconds?
    Well, not really. The 'exit speed' of the water should be relative to the bottle, but I suspect the author of the question intends to be kind to you here and wants you to use this as speed relative to the ground, as you have done.
    But what is the speed after all water has been emptied out? Don't forget that gravity has been acting on the bottle all this time.

    By the way, even allowing the given water exit speed as being relative to the ground, I disagree with the given answer. It's way too high.
    Last edited: Feb 22, 2015
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