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Finding the time and maximum height at highest point

  1. Sep 30, 2015 #1
    1. The problem statement, all variables and given/known data:

    A projectile is given initial velocity 80 m/s ( V0 ) at angle 60° above the horizontal. Find the time it takes to reach to the highest point and find the maximum height. (g = -10 m/s2)

    2. Relevant equations:

    1. [itex]y_{final} = y_{initial} + v_{y_{initial}}t + \frac {1}{2}gt^2[/itex]

    2. [itex]v_{final}^2 = v_{y_{intial}^2} + 2g(y_{final} - y_{intial})[/itex]


    3. The attempt at a solution:

    The first thing I did was to find the velocity of the y-component ( [itex]v_{y_{initial}})[/itex]:

    [itex]v_{y_{initial}} = (80 m/s)({\sin 60}^{\circ})[/itex]

    [itex]v_{y_{initial}} = 69 m/s[/itex]

    Since I know that [itex]v_{final}^2[/itex] is 0 m/s at the highest point, I can use Equation 2 to figure out the max height. With [itex]v_{y_{initial}^2}[/itex] known and [itex]t[/itex] unknown:

    [itex]v_{final}^2 = v_{y_{intial}^2} + 2g(y_{final} - y_{intial})[/itex]

    [itex]0 = v_{y_{intial}^2} + 2gy[/itex]

    [itex]y = \frac{-v_{y_{intial}^2}}{2g}[/itex]

    [itex]y = \frac{-(69 m/s)^2}{2(-10 m/s^2)}[/itex]

    [itex]y = 238 m[/itex]

    With the max height known, I can use Equation 1 to get the time to reach the max height:

    [itex]y_{final} = y_{initial} + v_{y_{initial}}t + \frac {1}{2}gt^2[/itex]

    [itex]238 m = (0m) + (69 m/s)t + \frac {1}{2}(-10m/s^2)t^2[/itex]

    [itex](5m/s^2)t^2 - (69 m/s)t + (238 m) = 0[/itex]

    From there, I used the quadratic formula where a = 5 m/s2, b = -69 m/s, and c = 238 m

    The results that I got were t = 175 s and t = 170 s, which doesn't seem right.

    Anyone care to point me to the right direction?
     
    Last edited: Sep 30, 2015
  2. jcsd
  3. Sep 30, 2015 #2

    gneill

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    Staff: Mentor

    A couple of things. First, keep some extra decimal places in your intermediate values that you will be using for further calculations. This will prevent rounding and truncation errors from creeping into latter calculations. Rounding for presentation purposes is fine, but use full precision in your calculations. So for example, your initial vertical velocity should be something like 69.282 m/s.

    Speaking of initial vertical velocity, you've shown the cosine function rather than sine function in its calculation. I presume this is a typo since you've show a valid value for the result.

    Something appears to have gone awry in your evaluation of the quadratic at the end. Try the calculation again.

    Note that if you happen to have some differential calculus under your belt, you can get at the time of the maximum height much more easily by just maximizing the vertical trajectory equation...
     
  4. Sep 30, 2015 #3
    I redid the quadratic calculation and got t = 69.1 s and t = 68.9 s, which still doesn't seem right.
     
  5. Sep 30, 2015 #4

    gneill

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    Staff: Mentor

    The answer should be less than 10 seconds. I guess you'll have to show your work in some detail so we can see what's happening.
     
  6. Sep 30, 2015 #5

    RUber

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    Homework Helper

    Don't forget the 2a in the denominator for the quadratic equation.
    As gneill said, your imperfect approximations are giving you imperfect answers.
     
  7. Sep 30, 2015 #6
    Ah, looks like I forgot to punch in a set of parenthesis in my calculator.

    t = 6.8 s

    I also just realized that I can easily calculate without having to use the quadratic formula the t with:

    [itex]v_{y_{final}} = v_{y_{intial}} -gt[/itex]
     
  8. Sep 30, 2015 #7

    haruspex

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    Gold Member
    2016 Award

    Better still, never plug in numbers until the end. In the specific case of finding the maximum height, you would have found ##s=\frac{v_0^2\sin^2(\theta)}{2g}##. Plugging in the angle, ##\sin^2(\frac{\pi}3)=\frac 34##, avoiding any approximations.
     
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