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Projectiles Honors

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data
    An arrow is fired from the top of a 60 m high cliff. It is fired at 80 m/s at a 30 degree angle above horizontal.

    a)Find the time of the arrow's flight from start to impact.
    b)Find the horizontal distance the arrow travels.
    c)Find the velocity of the arrow(magnitude and direction) when it strikes the ground.


    2. Relevant equations
    sine
    cosine
    x=vt
    t=(v-v0)/a
    3. The attempt at a solution

    a)
    I calculated the vertical and horizontal velocities.
    Horizontal: 80 cos 30= 69.8 m/s
    Vertical:80 sine 30= 40 m/s

    Then I calculated the time.

    v0=40 m/s
    a=-9.8 m/s^2
    v=0
    t=(v-v0)/a
    t=(0-40)/-9.8
    t=4.1

    This was considering the time from start to peak, so we double to get 8.2 s.

    b) x=vt
    x=69.28(58.2)
    x=4032 m.

    This seems conceivable, but not exactly right.

    c) How do I find the magnitude and direction? It is lost on me at the moment. If the arrow is *on* the ground, it's not moving, and so the velocity would be 0 m/s. But I am assuming the question means the velocity at the exact moment it hits the ground.

    We aren't taking air resistance into account.

    Thanks in advance! :smile:
     
  2. jcsd
  3. Dec 31, 2013 #2
    The time to the apex is half the total flight time only when the initial height is equal to the final height. Not the case here.
     
  4. Dec 31, 2013 #3
    Oh, okay! I didn't know that. So if I half the "total" time I gave, I'm good?
     
  5. Dec 31, 2013 #4
    Then you will only have time to the apex - the highest point of the trajectory - while you need time to the ground.
     
  6. Dec 31, 2013 #5
    Okay, that's what I thought. How do I calculate that? I'm not sure my class has gone over how how to find both legs of the "trip" when doubling the time doesn't work.
     
  7. Dec 31, 2013 #6
    Is the following equation familiar: ## y = y_0 + v_0 t - g \dfrac {t^2} 2 ##?
     
  8. Dec 31, 2013 #7
    Not at all. This is odd, I've watched all the lectures up to this point. Could we be overlooking a typo that makes the problem seem more complicated than it is?
     
  9. Dec 31, 2013 #8
    Hmm. You need to have some knowledge of uniformly accelerated motion to solve this problem. I notice that you know the equations for velocity in uniformly accelerated motion. What you need here is an equation for displacement in uniformly accelerated motion. Or knowledge of energy concepts, but that is probably more advanced.
     
  10. Dec 31, 2013 #9
    I do know some for displacement. Do you mind if I show you a similar problem my teacher solved in class? Maybe it will show something I missed.
     
  11. Dec 31, 2013 #10
    Sure, go ahead.
     
  12. Dec 31, 2013 #11
    Okay, this is the one I was referencing when I solved this problem.

    On the planet Mars, the gravitational acceleration is 3.7 m/s2. Suppose a rock is thrown upward
    at a speed of 20 m/s at an angle of 70 degrees above the horizontal. Find the time of the rock’s
    flight and the maximum height it reaches.

    v0x=20 m/s cos 70=6.84 m/s
    v0y=20 m/s sin 70=18.79 m/s

    Consider motion from start to peak.
    v=v0+at
    t=(v-v0)/a
    t=(0-18.79 m/s)/-3.7 m/s^2
    t=5.08 s
    Total time=10.16 s

    y=y0+v0t+1/2a*(t^2)
    =0+18.79 m/s(5.08s)+1/2(-3.7 m/s^2)(5.08s)^2
    =95.45-47.74
    =47.71 m
    =48 m

    Thanks! :smile:
     
  13. Dec 31, 2013 #12
    This is the same equation I mentioned in #6. Which you said you did not know :)

    This equation is key to the problem. Notice you have ##y##, ##y_0##, ##v_0## and ##a = -g##. The only unknown is ##t##. Which is what part (a) is about.
     
  14. Dec 31, 2013 #13
    Oh! I didn't notice, sorry. XD The (-g t^2)/2 looked unfamiliar. I need to use *that* equation to solve for time, instead of the one I tried?
     
  15. Jan 1, 2014 #14
    Yes, you could use that equation to solve this problem.
     
  16. Jan 1, 2014 #15
    Okay. Let's see how my remembrance of pesky algebra concepts holds up.

    y=y0+v0t+1/2a*(t^2)

    If I multiply both sides by 2 it will get rid of the 1/2.

    2y=y0+v0t+a*t^2

    Let's plug the numbers in to make it easier.

    Okay, I think I can solve for t, but I'm a bit confused on some parts. y0=60 m, correct? We're calculating the vertical? So what is y?

    Thanks! :smile:
     
  17. Jan 1, 2014 #16

    SteamKing

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    Science Advisor
    Homework Helper

    You want to take another go at multiplying both sides of an equation by 2? Those 'pesky algebra concepts' should be learned thoroughly if you expect to get good grades in any technical courses.
     
  18. Jan 1, 2014 #17
    Would you multiply *all* the terms by two? I know the concepts, I just tend to be better at them when I put the numbers I know in. That's why I asked what "y" is in this case. This problem will be a good refresher.:smile:
     
  19. Jan 1, 2014 #18
    You should not be asking those questions. I suggest that you go over the chapter in your textbook (or lecture/course notes) that explains that equation with all those symbols. It seems that you are guessing at it, and that is no good.
     
  20. Jan 1, 2014 #19
    I already did. I know what the symbols mean, but I wasn't sure what y final is because I couldn't seem to find the part that told us where the arrow landed. Now I see that in part c) it says it strikes the ground, so if we let down be positive y0=0 and y=60.
     
  21. Jan 1, 2014 #20

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, when you multiply both sides of an equation by a constant, 'all' terms in a sum get multiplied by the constant. That's not even algebra; it comes from basic arithmetic:

    a*(b + c) = a*b + a*c
     
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