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Math Help - Physics: Magnitude of initial velocity

  1. #1
    Newbie cheet0face's Avatar
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    Physics: Magnitude of initial velocity

    Problem:
    A frog makes a mighty leap, leaving the ground at an angle of 45 degrees and traveling a horizontal distance of 1.47 (m) in 0.55 (s).

    a)What is vx?
    b)What is vyi?
    c)What is the magnitude of the initial velocity, at an angle of 45 degrees?

    Now I solved question b by drawing a little figure and finding vyi, and used that to find time and then used that to find vx. But im not quite sure how to find the magnitude?

    Δy=?
    vyi=2.67 (m/s)
    vyf=?
    a=-9.8 (m/s^2)
    t=.55 (s)

    Δx=1.47 (m)
    vx=2.76 (m/s)
    t=.55 (s)

    Thanks in advance
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  2. #2
    Senior Member
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    a)What is vx?
    I got 2.67 instead of 2.76, but I am guesssing that that's just a typo.

    b)What is vyi?
    I got a different answer to yours. We have
    a = -9.8
    Δy=0
    t=.55 (s)
    vyi=?
    Since we don't care about final velocity, use the equation \delta y = v_{yi}t+0.5at^2 and solve for  v_{yi}

    c)What is the magnitude of the initial velocity, at an angle of 45 degrees?
    magnitudes of vectors can be found using triangles like this one. You know \theta = 45^o and v_{yi}. You want |V_i|Physics: Magnitude of initial velocity-magn.jpg
    Last edited by badgerigar; September 18th 2008 at 09:51 PM. Reason: wrong formula
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  3. #3
    Newbie cheet0face's Avatar
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    Thanks!
    That helped a lot

    I got the magnitude as 3.77 (m/s). Is this correct?

    Another question I have is on a different problem similar to this one and they want to know the direction, and I have no idea how to find that?
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  4. #4
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    I got the magnitude as 3.77 (m/s). Is this correct?
    No. But I think it was my fault: the formula for viy I posted before was wrong and has now been fixed. You should get 2.95 for viy and about 3.8113055505954911562055111 for the magnitude

    Finding theta in the triangle above gives direction
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