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Initial Velocity of Projectile Motion, given displacement and initial angle?

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A grasshopper jumps 1.00 meters from rest, with an initial velocity at a 45.0° angle with respect to the horizontal. Find (a) the initial speed of the grasshopper and (b) the maximum height reached.


    2. Relevant equations
    vx = v0xt = v0cos θ0

    vy = v0 sin θ0 - gt


    3. The attempt at a solution
    I was unsure what to do, so I resolved for the components, resulting in

    v0x = v0 cos 45°

    v0y = v0 sin 45°

    I apologize for my cluelessness - this my first semester in physics (at all), and I am having trouble finding where to go about this problem. Thank you in advance.
     
    Last edited: Oct 23, 2012
  2. jcsd
  3. Oct 23, 2012 #2
    is that 1.00 supposed to be in meters? Is the grasshopper jumping 1.00 meter forward across the ground?

    also, when the grasshopper reaches the peak of its arc through the air, what is the magnitude of the *vertical component* of the velocity? (this is a question that is trying to push you in the right path, the first one is clearing up some confusion on what the problem is asking)
     
  4. Oct 23, 2012 #3
    Yes, meters - sorry. I have edited the above.
     
  5. Oct 23, 2012 #4
    The velocity of the peak of the arc would be 0. I tried to look for a way to use this to find the initial speed but I'm stuck, is there a way I can use this information to do that?
     
  6. Oct 23, 2012 #5
    well so you can solve for the t it takes to get to that point

    and so then if that's the amount of time it takes to get to the *peak*, how long does it take the grasshopper to go the full distance?
     
  7. Oct 23, 2012 #6
    So far, I've solved using Vy sin θ0 - gt
    to get

    tmax = v0sin 45°/9.8

    ttotal = 2(v0 sin 45°/9.8)

    Without the initial velocity, I'm not sure how (if I can) to simplify any further, or solve for it..
     
  8. Oct 23, 2012 #7
    45 degrees is right in the middle of completely horizontal and completely vertical. You might feel like the horizontal and vertical components should be the same, and indeed they are. cos45=sin45 Might as well take Vox and call it Voy

    X = Voyt ,where X is the horizontal distance traveled.

    Vyf = Voy -gt

    However Vyf is right before hitting the ground... it must have equal magnitude but opposite direction of Voy so:

    -Vyo = Voy -gt

    So you have 2 equations, two unknowns and can solve.
     
    Last edited: Oct 23, 2012
  9. Oct 23, 2012 #8
    so if you know that the time taken for the grasshopper to end its jump is ttotal = 2(v0 sin 45°/9.8)

    and that it went 1 meter

    and you know that

    x = v0,xt = v0cos(θ)t

    what could you do with that
     
  10. Oct 23, 2012 #9
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