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Maximum height of a spring toy launched at an angle

  1. Dec 6, 2017 at 7:24 PM #1
    1. The problem statement, all variables and given/known data
    A spring toy is launched from the ground at 2.3 m/s at an angle of 78° to the ground. What is the maximum height reached by the spring toy?

    2. Relevant equations
    Δt = 2(V1)Sinθ/(acceleration)
    ΔD = V1Δt + 1/2(acceleration)(Δt^2)

    3. The attempt at a solution
    First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)
    Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)
    Δt = 0.46s


    Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)
    ΔD = (2.3m/s)0.46s + 1/2(9.8 m/s)(0.46^2)
    ΔD = 1.058 + 1.037
    ΔD = 2.095m

    Is this the right answer? Is this the right equation or should I be using ΔDv = 1/2(acceleration)Δt^2 ?Somehow this doesn't seem right because I don't know when the toy is at the highest point. If I knew the time of its highest point then I could find the vertical distance using that time, that is how I have been taught.

    One more thing, should I be saying [up] is positive and using - 9.8 m/s[up] in my above equations?
    e.g. ΔD = (2.3m/s[up])0.46s + 1/2(-9.8 m/s[up])(0.46^2)

    Thanks for any help
     
  2. jcsd
  3. Dec 6, 2017 at 8:46 PM #2

    gneill

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    Staff: Mentor

    Hi Cat 2,

    Welcome to Physics Forums.

    You might notice that I've changed the title of your thread to make it descriptive of the problem being posed. You want to use highly descriptive titles in order to grab the attention of the Homework Helpers with the right skill set :wink:

    In your attempt, what time is being calculated by the formula that you've employed? Are you sure that the maximum height occurs at that time?
     
  4. Dec 7, 2017 at 8:21 AM #3
    Thanks gneill,

    OK, I'm not sure that us where the maximum height occurs. I don't have a formula for finding maximum height nor have I had a question that asked this before. I said above......
    That is what I have been taught to do.

    The formula I used, found the time of the flight, 0.46s. Should I be saying that the highest the toy goes would be half of the time of the flight, 0.23s? (Half of a parabola is its maximum point and vertex.)
     
  5. Dec 7, 2017 at 8:31 AM #4

    gneill

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    Yes!

    Alternatively, you could derive an expression for the time of maximum height by writing the equation of motion for the y-component and using a bit of calculus to find when the y-value is maximized.
     
  6. Dec 7, 2017 at 8:59 AM #5
    OK, so would my answer look like this then?

    First I solved for time using this equation: Δt = 2(V1)Sinθ/(acceleration)
    Δt = 2(2.3 m/s) Sin78 / (9.8 m/s)
    Δt = 0.46s

    0.46s/0.5 = 0.23s - this is half of the total time of my parabola, therefore it will be the highest point.

    Then I solved for distance using this equation: ΔD = V1Δt + 1/2(acceleration)(Δt^2)
    ΔD = (2.3m/s)0.23s + 1/2(9.8 m/s)(0.23s^2)
    ΔD = 0.529 + 0.259
    ΔD = 0.788 meters
     
  7. Dec 7, 2017 at 9:41 AM #6
    Not quite. Two things here,

    Thinking generally about the physics here, do you expect that the acceleration due to gravity will have a positive contribution to the max upward distance or negative one? Is that in agreement with your calculations? What are the directions here?

    What is the 2.3sin(78) in the equation for t? Do you understand why that is in the equation for t? Should it appear in the equation for D?
     
  8. Dec 7, 2017 at 10:02 AM #7
    Thanks for helping RedDelicious!
    My acceleration should've been -9.8 m/s^2 [up], and the speed, 2.3m/s [up].

    Also (2.3m/s[up])0.23s should have been (2.3m/s[up])(sine78)0.23s

    ΔD = V1Δt + 1/2(acceleration)(Δt^2)
    ΔD = (2.3m/s[up])(sine78)0.23s + 1/2(-9.8 m/s^2[up])(0.23s^2)
    ΔD = 0.517 - 0.259
    ΔD = 0.258 meters

    How's that?
     
  9. Dec 7, 2017 at 12:49 PM #8
    Did I do that correctly now?
     
  10. Dec 7, 2017 at 2:10 PM #9

    gneill

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    Good attem
    Good!

    A couple of technical points about the presentation of your equations. If you start by defining your coordinate system, you don't have to label each value with [up] (or [down]). The appropriate direction is implied by the sign of the value once the coordinate system is defined. Thus:

    Assuming up is the direction of the positive y-axis:

    ΔD = (2.3m/s)(sine78)0.23s + 1/2(-9.8 m/s^2)(0.23s)^2

    Note also that I moved the final parenthesis so that the entire time quantity, including its units, are enclosed for the squaring operation. Otherwise one might misinterpret things, supposing that only the units were to be squared.
     
  11. Dec 8, 2017 at 8:38 AM #10
    Alright, I think I got it all! Thanks a lot for all your help gneill!
     
  12. Dec 8, 2017 at 2:40 PM #11

    haruspex

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    I recommend you delete this equation from your notes. The trouble with long lists of memorised equations is that you forget which apply when. Better to keep a short list of the most general equations.
    The one you quote is for the specific case of time to return to the same height.

    There are five "SUVAT" equations that cover all the constant acceleration problems you will encounter. The acronym represents the five variables, s=displacement, u= initial velocity, v=final velocity, a=acceleration, t=time. Each equation omits one variable and relates the other four; identify which three you know (or connect to other facts), which one is to be determined, and select the equation involving those four.

    See e.g. https://en.m.wikipedia.org/wiki/Equations_of_motion

    In the present case, you initial vertical velocity, vertical acceleration, and final vertical velocity (0 at highest point) and you wish to find the time.
     
  13. Dec 9, 2017 at 10:01 AM #12
    Wait, as I am fixing this problem and adding corrections (-9.8 m/s [up] and 2.3 m/s [up]), I find that if I apply this to the second equation I get a positive answer........
    ΔD = V1Δt + 1/2(acceleration)(Δt^2)
    ΔD = (2.3m/s[up])(sine78)0.23s + 1/2(-9.8 m/s^2[up])(0.23s^2)
    ΔD = 0.517 - 0.259
    ΔD = 0.258 meters

    But if I apply this to the first equation and then the second equation, I get a negative answer........
    Δt = 2(V1)Sinθ/(acceleration)
    Δt = 2(2.3 m/s) Sin78 / (-9.8 m/s)
    Δt = -0.46s

    ΔD = V1Δt + 1/2(acceleration)(Δt^2)
    ΔD = (2.3m/s)(sine78)-0.23s + 1/2(-9.8 m/s^2)(-0.23s)^2
    ΔD = - 0.517 - 0.259
    ΔD = - 0.776 meters

    Do I a) did something wrong. (I don't think so.)
    b) ignore (remove) the negative sign and this is my answer. (Maybe? But that doesn't make sense because the - has a purpose. )
    c) not use (-9.8 m/s [up] in the first equation. ( Maybe, because I am just finding time and not distance so it doesn't matter?)

    What do I do to not get a negative answer? use c) ?
    Please Help! I really need to get this!
     
  14. Dec 9, 2017 at 10:58 AM #13

    gneill

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    You need to look at the derivation of your formula:
    in order to make sure that you understand where and how to apply it. Applying formulas without really understanding where they come from is not recommended. It can lead to unexpected errors.

    Can you derive that equation from the general motion equation ##s = v\;t + \frac{1}{2} a t^2##? I think you'll spot the problem you had if you do.
     
  15. Dec 9, 2017 at 12:58 PM #14
    No, unfortunately I completely don't get what deriving the equation from the general motion equation has to do with the problem. Could you help me out a little more? Does it have to do with what your finding e.g. if your finding displacement then you need directions, but if your finding time, not?
     
  16. Dec 9, 2017 at 1:02 PM #15

    gneill

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    You are blindly applying a formula with unwarranted assumptions about the variables in it. Hence, a sign error has occurred giving you an incorrect (negative) value for the time.

    Deriving the equation yourself would give you insight into the issue, and further, it would be one less equation to memorize for your tests and exams: you could easily derive it in a matter of moments from the basic equation of motion, and be certain of what the variables mean and what signs are appropriate.
     
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