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Projectile motion: grasshopper

  1. Jan 26, 2008 #1
    A grasshopper leaps into the air from the edge of a vertical cliff, as shown in the figure below. a.) Use information from the figure to find the initial speed, in m/s, of the grasshopper. b.) Use information from the figure to find the height, in m, of the cliff.
    The grasshopper jumps at an angle of 50, reaching a maximum height of 6.74 cm above the cliff, and hitting the ground below the cliff 1.06 m away.


    So I think the relevant equations for it at Vx=Vocos([tex]\theta[/tex]) and Vy=Vosin([tex]\theta[/tex]) for the x and y components, honestly I'm really lost, this is my first physics course. Also, I think xfinal-xinitial=Vxt+1/2gt^2, and since the acceleration in the x-direction is zero, it's just xfinal-xinitial=Vxt. And then yfinal-yinitial=Vyt+1/2gt^2.

    I'm not sure how to find the initial speed, because I'm not sure how to go about finding t. I set 1.06=Vocos(50)t equal to .0674=Vosin(50)t+1/2(9.8)t^2 but since I don't know what Vo is or the time, I'm not sure how to solve for either. Then once I find what either is, I have no clue how to find the height of the cliff. I'd appreciate any help that could send me in the right direction. :)
     
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  3. Jan 26, 2008 #2

    Hootenanny

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    What do you know about the particle when it is at it's maximum height? [Hint: What is it's velocity and what is it's displacement above the cliff?]
     
  4. Jan 26, 2008 #3
    At it's maximum height, isn't the velocity zero? And at that point it's .0674 m above the cliff, but I'm just not sure where to put that information so I can find the initial velocity and the height of the cliff.
     
  5. Jan 26, 2008 #4

    Hootenanny

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    Indeed, you are correct. Perhaps this would be helpful;

    [tex]v^2 = v_0^2+2a(s_2-s_1)[/tex]
     
  6. Jan 26, 2008 #5
    so then, vo would be 0^2? So would the velocity be V=[0+2*9.8*(.0674-0)]^.5=1.15m/s?
     
  7. Jan 26, 2008 #6

    Hootenanny

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    Here, v0 is the initial speed, v is the speed at a given displacement. v0 is what you need to solve for.
     
  8. Jan 26, 2008 #7
    ok, so i had plugged 0 in the wrong spot? so it should be 0^2=vo^2+2*9.8*(.0674-0) but if i just move where i put the zero, it would be (-1.32)^.5, which can't be done unless you have imaginary numbers, and even if you could, wouldn't it still equal 1.15? So i think i did something wrong. should I be taking into account that the grasshopper ended up 1.06 m away from the initial place it jumped?
     
  9. Jan 26, 2008 #8

    Hootenanny

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    That would be the case if gravity caused mass to accelerate upwards, but it doesn't. Gravity acts downwards and so in our frame, the acceleration due to gravity is negative, a = -9.8 m/s2. Does that make sense?
     
  10. Jan 26, 2008 #9
    Yea, it makes sense as to why gravity is negative. So, then is the answer 1.15 m/s? or do i still have to put that into another equation?
     
  11. Jan 26, 2008 #10

    Hootenanny

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    Nope, 1.15m/s is the intial vertical velocity (assuming that you did the arithmetic correctly). So, armed with this information, can you calculate the magnitude of the intial velocity and hence the horizontal component of the intial velocity?
     
  12. Jan 26, 2008 #11
    so would it be 1.15/cos(50), which comes out to 1.78?
     
  13. Jan 26, 2008 #12

    Hootenanny

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    Careful, 1.15 is the vertical component, so to find the magnitude of the intial speed you shoud evaluate 1.15/sin(50). Do you follow?
     
  14. Jan 26, 2008 #13
    ok, so then it should be 1.5? and then i would divide that by cos(50)?
     
  15. Jan 26, 2008 #14

    Hootenanny

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    Okay, I think we need to refresh here. Given a speed v the horizontal component (vx) and the vertical component (vy) is given by;

    [tex]v_x = v\cos\theta \Rightarrow v = \frac{v_x}{\cos\theta}[/tex]

    [tex]v_y = v\sin\theta \Rightarrow v = \frac{v_y}{\sin\theta}[/tex]

    Where [itex]\theta[/itex] is the angle above the horizontal.

    P.S. I'm not checking your arithmetic here (I haven't got a calculator handy).
     
  16. Jan 26, 2008 #15
    ok so for the magnitude of the initial speed, it will be 1.15/sin(50) and i got 1.5 for that, and that would be my initial speed? So, once i have that, how do i use it to find the height of the cliff? OH now i can solve for t, right?
     
  17. Jan 26, 2008 #16

    Hootenanny

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    Correct.
    Not quite, you still don't have enough information. What you need to do is find an expression for time in terms of your intial horizontal speed and horizontal displacement. Then, you can write an expression for the vertical displacement in terms of time. You now have two simultaneous equations for time which you can solve.

    I wish I had more time for a more detailed explantion, but unfortunatly, I'm very very very tired and its very late/early :zzz:.
     
  18. Jan 26, 2008 #17
    alright, well thank you so much for your help, i think i'm gonna try an keep working on it
     
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