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Projectile Motion Clarification/Help

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello everyone, first time poster, long time lurker here. I'm currently enrolled in an online grade 11 physics course and don't have immediate access to a teacher, nor do they seem as helpful and friendly as PF! My algebra skills are not the best, so I may need a little clarification during this process.

    Anyways, the problem presented below has me a little confused, because in my book, I'm supposed to let displacement equal 0 and find time, but there are two instances of time in the equation. What I'm confused about is that the book shows that the first time ( [itex] \Delta t [/itex] ) just vanishes and doesn't explain why.

    I'm assuming it's being brought to the other side of the equation by dividing by 0. There comes another problem, at least in my book where the second time ( [itex] \Delta t^2 [/itex] ) is being factored out, which I don't understand because why wouldn't you just isolate it and then bring the square over, which gives the answer 5.06 s, whereas, in the book it says the answer to be 0 and 6.2 s?

    Am I completely wrong? Is this a typo in my book? Am I insane?


    A golf ball is hit with an initial velocity of 40 m/s at an angle of 50 degrees.

    a.) How long is the ball in the air?




    2. Relevant equations

    [itex]\vec{v}_iv=\vec{v}_i sin \theta [/itex]

    [itex]\Delta \vec{d}=\vec{v}_iv \Delta t +\frac {1}{2} \vec{a} \Delta t^2 [/itex]

    3. The attempt at a solution

    [itex]\vec{v}_iv=(40 m/s) sin 50° [/itex]
    [itex] \vec{v}_iv=30.6 m/s [up] [/itex]


    [itex] 0=(30.6 m/s) \Delta t - 4.9 m/s^2 \Delta t^2 [/itex]
    [itex] \Delta t= √ 25.7 [/itex]

    [itex] \Delta t= 5.06 s [/itex]
     
  2. jcsd
  3. Nov 19, 2013 #2

    haruspex

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    Ignoring air resistance, projectiles follow a parabolic arc. How many times will it be at any given altitude (below the peak)? In the current question, what is the significance of the other solution?
     
  4. Nov 20, 2013 #3
    I'm not entirely sure what you mean, but I'm trying to find how long the ball was in the air for. Are my calculations incorrect?

    Sorry if I'm coming off confusing, I'm really snagged on this problem, mostly by the algebra but also if my answer differing from the textbook's answer.
     
  5. Nov 20, 2013 #4

    haruspex

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    Your calculation was fine to here:
    [itex] 0=(30.6 m/s) \Delta t - 4.9 m/s^2 \Delta t^2 [/itex]
    But I don't understand what you did next. You should have got two solutions for Δt, neither involving a square root. One of the solutions you can discard - what is it, and why?
     
  6. Nov 20, 2013 #5
    This seems to be where I'm going wrong, I'm not fully sure, algebraicly speaking, how to go about isolating time in this equation. My book is very vague when showing how it's supposed to be isolated.
     
  7. Nov 20, 2013 #6

    haruspex

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    You have the quadratic 0 = ut + at2.
    There are two ways to proceed with a quadratic. You can either use the standard solution formula http://www.purplemath.com/modules/solvquad4.htm, or spot factors. Can you see how to factorise ut + at2?
     
  8. Nov 20, 2013 #7
    I've used the quadratic equation in a previous math course but so far in my physics course there has been no use or mention of it. I can see how it would give me the roots of the parabola but I feel like there's another way for me to find the information I need. Like I said, I'm not the greatest with algebra, if you could give me a little more guidance, that would be wonderful. Thank you for being so patient.
     
  9. Nov 20, 2013 #8

    haruspex

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    ut + at2 has a factor t. Factorise that out. You now have two factors whose product is zero. What does that tell you about the two factors?
     
  10. Nov 21, 2013 #9
    I don't understand how they factor out though, there's 2 of them.
     
  11. Nov 21, 2013 #10

    haruspex

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    Like so: ut + at2 = t(u+at)
     
  12. Nov 21, 2013 #11
    So at this point the equation looks like: t=t(u+at) ?
     
  13. Nov 21, 2013 #12

    haruspex

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    No, you had 0 = ut + at2/2 (I left out the /2 before). Factorise the RHS like I did. What do you get?
     
    Last edited: Nov 22, 2013
  14. Nov 21, 2013 #13
    I don't get it! Frustration befalls me!

    My roadblock is that there's a time variable being multiplied by velocity and also a squared time variable being multiplied by acceleration. Now I understand that the squared time variable can be factored like you were showing, but I'm confused on what happens to he third 't' in the equation.
     
  15. Nov 22, 2013 #14

    haruspex

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    ..and I don't understand what you don't understand :confused:
    You have 0 = ut + at2/2. Both terms on the right have a factor t, so you can factor that out:
    0 = (u + at/2)t
    Do you understand that those two equations are the same?
    So now you have two things, t and u + at/2, which when multiplied together give 0. If multiplying two numbers together gives 0, what can you say about the numbers?
     
  16. Nov 22, 2013 #15
    I see what you're saying now, I was getting hung up on the two time variables but I see now they are 2 separate answers, one wi equal zero and the other the total time in the air.

    Thank you kind sir for walking me through the process. You are a gentleman and a scholar, good day to you!
     
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