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Homework Help: Need help solving a tricky projectile motion problem

  1. Feb 28, 2016 #1
    1. The problem statement, all variables and given/known data
    A fountain shoots water from a fountain that is 3.5cm from the ground. The water reaches a maximum of height of 11cm from the ground (or 7.5cm from the height of the fountain). The range of the water is 15cm from the fountain once it hits the ground. Find:
    a) The time the water spends in the air
    b) The initial speed of the water
    c) The initial angle at which the water is shot

    2. Relevant equations
    eq.1) Vxi = Vicos(Θ)
    eq.2) Vyi = Visin(Θ)
    eq.3) Xf = Xi + Vxit
    eq.4) Yf = Yi + 0.5(Vyi + Vyf)t
    eq.5) Yf = Yi + Vyit - 0.5gt2
    *possibly a few more

    3. The attempt at a solution
    I have been attempting this one for a while now, with no luck. I suspect it involves a system of equations. I have tried first using equations 4 and 5 to solve for t, but I do not think it gave me the right value for t, since I then tried using t in equation 4 to solve for Vyi, but I also do not think that the value of that is right. I tried from there to use more equations, but the angle ended up being undefined or out of the domain of 0<x<90. I am very lost, any tips for at least where to start would be helpful. My professor did leave some hints:
    -Solve for the time to its peak, and then for the time back down, and add them.
    -Solve for Viy given the information regarding max. height.
    But I still can't figure out where to even begin. All I know is that the y-velocity is 0 at the top, Xf=15, Xi=0, Yf=0, Yi=3.5, range or ΔX = 15, and ΔY = 11
  2. jcsd
  3. Feb 28, 2016 #2


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    You've made a mistake in assigning a value to one of your variables. Yf ≠ 0. How high does the water project from the fountain?

    BTW, it's better and more informative to show your actual calculations rather than describe what you did verbally. By examining your calculations, we might be able to determine where you are going wrong.
  4. Feb 28, 2016 #3
    This seems fairly simple. For problems involving projectiles, taking the motion along each axis individually helps greatly. Let's look at part (a) and (b), you know delta(y) and the acceleration along that axis ( is it g? ). Using these both can you find Vy and the time taken to reach the top? After that, in the same direction you can find the time taken for the water to reach the ground ( you know the max. height). So this was all in the y direction.
    For part (c), you found out Vy in the previous part, so if you could calculate the velocity in the x direction Vx ( is it constant?), you can get the angle. No?
  5. Feb 28, 2016 #4
    Either you can find the time easily, by just going to the point that the water is at its maximum fall and treat it as a free falling object. But this is the time only for it going down. How do you think we should get the air time?

    Or you can mess around with the equations to get 2 equation solve them and you will end up with the angle then the velocity then you can calculate the airtime (a bit upside down XD)
    You will find that sin(theta)2 = sin(2theta)
  6. Feb 28, 2016 #5


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    Just to add to the info, please remember that the "time of flight" is dictated ONLY by the motion in the vertical direction. In other words, if you find the vertical component of the initial velocity, you will get the same time of flight if you simply shoot the water vertically upwards with that initial velocity and then let it fall to the ground.

    That should address part (a) of your question, and you might gain an insight on how to do the rest after that.

  7. Feb 28, 2016 #6
    Sorry for the delay guys, had to go to work. So I do understand that the time the projectile is in the air only relies on it's y-component of velocity. So for solving part (a), I believe I need to use the equation
    Vyf = Vyi - gt ==> 0 = Vyi - (980)t ==> Vyi = 980t
    So now I have two unknowns, so I guess I could use
    Yf = Yi + Vyit - 0.5gt2 ==> 11 = 3.5 + 980t2 - 490t2
    When I solve for t, my calculator gives me about 0.12. Did I do my system of equations correctly? I now need to find the time it takes for it to descend. I saw that someone said Yf ≠ 0. Is it perhaps -3.5 then? I assumed it was zero because I took the ground to be 0, so if the Yi is the height of the fountain (3.5) wouldn't Yf = 0 since the water falls to the ground?
  8. Feb 28, 2016 #7


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    Why can't you use this kinematical equation? vy2 = vy02 + 2ay

    You know the maximum height, and you know that at that height, the y-component of the velocity is zero.

  9. Feb 28, 2016 #8
    Oh okay, I must have missed that. So I got 121.244 for Vyi. So now I will use that in eq.4 to solve for time. One question I still have is what is the Yf? I know Yi is 3.5cm. So Yi 0 (the ground)?
  10. Feb 28, 2016 #9


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  11. Feb 28, 2016 #10
    The units are in cm. It seems a little high, but I can't see any error in my calculation. I proceeded to use eq. 5 to find the time:
    0 = 3.5 + 121.24t - 0.5(980)t2 = 0.27s
    I then used eq. 3 to find Vxi:
    15 = 0 + Vxi(.27) = 55.56cm
    I then converted the x and y components into the actual velocity vector:
    √55.562 + 121.242 = 133.36 cm/s
    Finally, I used tan-1(121.24/55.56) = 65.38°
    I feel like the speed is kind of high, I don't know why. But do my calculations look correct?
  12. Feb 28, 2016 #11


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    Looks about right, but you should not take the output of a calculation step to more sig figs than some of the inputs. Rounding off the time to 0.27s is not a good idea.
    You could check the answer by working it backwards: given that speed and angle, should it reach that height and that distance?
  13. Feb 28, 2016 #12
    Oh yes, you are right. Thank you for the help, along with everyone else who contributed.
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