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Projectile Motion and speed

  1. May 21, 2008 #1
    [SOLVED] Projectile Motion Problem

    1. The problem statement, all variables and given/known data
    A frog jumps from a log and lands on the ground a distance of 2.80 m away. The frog is airborne for 2.5 seconds, and leaves the log at an angle of 40 degrees. (a) What was the frog's speed (vo) just after it jumped? (b) How high above the ground did the frog jump from?

    2. Relevant equations
    The equations I used:
    [tex]
    y = y_0 + v_0_yt - (1/2)g(t^2)
    [/tex]
    [tex]
    x = x_0 + v_0_xt
    [/tex]

    3. The attempt at a solution

    First I found that
    [tex]
    v_0_y = v_0sin(40)
    [/tex]
    and
    [tex]
    v_0_x = v_0cos(40)
    [/tex]
    Then,
    [tex]
    x = x_0 + v_0_xt
    [/tex]
    [tex]
    2.8 = 0 + (v_0*cos(40))(2.5)
    [/tex]
    [tex]
    v_0 = 1.4621 m/s
    [/tex]
    Now,
    [tex]
    v_0_y = v_0sin(40)
    [/tex]
    [tex]
    v_0_y = 1.4621sin(40)
    [/tex]
    [tex]
    v_0_y = 0.9398 m/s
    [/tex]
    Using the first equation,
    [tex]
    y = y_0 + v_0_yt - (1/2)g(t^2)
    [/tex]
    [tex]
    0 = y_0 + (0.9398)(2.5) - (1/2)9.8(2.5^2)
    [/tex]
    [tex]
    y_0 = 28.28 m
    [/tex]

    ...Now I really don't think that a frog would be jumping off a 30 meter log. Any ideas at where I'm going wrong?
     
    Last edited: May 21, 2008
  2. jcsd
  3. May 21, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    That looks to be correct.

    You could have simply also said that since the horizontal component of velocity stays the same. Distance=Speed*Time.

    For the second part. 2.5 seconds is the time for the frog to complete its parabolic motion. Which means that t=2.5 isn't the time at maximum height.
     
  4. May 21, 2008 #3
    Hmmm, I still seem to be getting the same answer.

    Using:
    [tex]
    v_y = v_0_y - gt
    [/tex]
    [tex]
    0 = 0.9398 - (9.8)t
    [/tex]

    I found that it takes roughly 0.06 seconds for the frog to reach the highest point in the jump leaving 2.4 seconds of freefall.

    [tex]
    y = y_0 + v_0_yt - (1/2)gt^2
    [/tex]
    [tex]
    0 = y_0 + (1/2)(9.8)(2.4^2)
    [/tex]

    Which still gives me a height of about 28m.
     
  5. May 21, 2008 #4
    Whoops...that last post calculated the maximum height reached, not the height at which the little bugger started from. I'm not really sure why I couldn't use 2.5s as the time like I did before.
     
  6. May 22, 2008 #5

    alphysicist

    User Avatar
    Homework Helper

    Hi dano404,

    The work in your original post looked okay to me. Was the large height not sounding right? I do understand that a log 30 meters in the air sounds strange, but the numbers they give seem consistent with your answer. Think about the time of flight and what it does to the y velocity. If the frog had just dropped off, after 1 second it would be going about 10 m/s, after 2 seconds it would be travelling about 20m/s, etc., so travelling about 30 m in 2.5 seconds sounds reasonable, I think. (The frog is travelling upwards at the beginning in this problem, but the initial y-velocity is relatively small so these approximate values are still close.)

    The poor frog may not enjoy the landing, but collisions are another problem.
     
  7. May 22, 2008 #6
    My professor had accidentally made a typo when writing the problem (on a test from a previous course). He typically tends to make things very realistic which is why I was surprised by the result. Thank you for the responses. I just took my exam, definitely aced it.
     
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