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Homework Help: Help, KE problem, not sure where to start

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Judy looks from her 20m high balcony to a swimming pool below, not directly below, but 3m out from the bottom of the building. If the pool is 6m wide. (a) How fast would she have to jump horizontally to land in the middle of the pool? (b) Judy's mass is 60 kg, what will be her increase in KE (compared to when she left the balcony) as she hits the water?

    2. Relevant equations
    horizontal velocity must play a factor
    possibly d=1/2gt^2

    3. The attempt at a solution
    The center of the pool is 6m out from the balcony (3m out plus 3m to center). I am very unsure of what to do beyond this. This is what I have started w...
    mgh= 1/2m(gt)2 (since gt=v)?
    divide by 2 then g's & m's cancel out ?
    leaving 2h=t^2
    40 square root = 6.324s which is the time?? I am REALLY digging here!! help!
    I'm not sure the even helps get me closer to am answer.
  2. jcsd
  3. Nov 11, 2008 #2
    remember GPE(balcony) + KE(balcony) = GPE(pool) + KE(pool) for the KE portion.
    As for the part you've done, knowing the time to the bottom is critical but I'm not sure what you've done. mgh in this is 60kg*9.8m/s*20m and that's your gravitational potential energy. (GPE) you've equated it to the KE equation which will help you solve for b only.
    to solve for a you must use y = Vi(t) + g(t^2) and solve for t. getting this will let you determine answer a because you'll know how many seconds she'll have to travel 6m.
    Hope this helps.
  4. Nov 11, 2008 #3
    Where is the 60kg from? is it 20m*3m?
  5. Nov 11, 2008 #4
    ...b) Judy's mass is 60kg...
  6. Nov 11, 2008 #5
    Is it ok to "borrow" that info from the second part of the question? I don't know the rules of physics.
  7. Nov 11, 2008 #6
    It part of the given of the question. Were it to be an entirely different question you couldn't but it's part b to the question, and even identifies it as being Judy.
  8. Nov 11, 2008 #7
    oh! Sorry. I'm new at physics ;)
    to solve for t
    y = Vi(t) + g(t^2) and solve for t.
    Since she is starting from rest (assumably) Vi is 0(t) + 9.8(t^2), then...I'm not sure. Could you give me a nudge.
  9. Nov 11, 2008 #8
    11,760 = 30 v^2 both sides /30
    v=square root of 392 = 19.79 ?
  10. Nov 11, 2008 #9
    Is the answer 19.79 m/s the answer for a? or am I solving for nothing? haha
  11. Nov 11, 2008 #10
    no biggie about being new to it so am I, I've got two questions up that I'm totally lost on.

    well so y = 20m = 9.8(t^2) once you solve this for t you can solve to your initial velocity off the balcony by taking your 6m/t that can be used to give you your initial KE off the balcony. but I wouldn't use that to solve for the KE(pool), I would only use it to determine the difference. I would set KE(balcony) to 0, and GPE(pool) to 0.
  12. Nov 11, 2008 #11
    Oh no, not the answer for a! The would be the velocity she hits the pool at. I think to solve for b you need to leave it in Joules, and then figure out a's answer, convert it to Joules and subtract them to get b's answer.
  13. Nov 11, 2008 #12
    I am not sure how to get the answer for (a). I though I was on the right track. The gpe on the balcony is 11,760, the KE on the balcony is 0. Then, she takes off... Her GPE decreases and her KE increases until she hits the water. Once she hits the water her GPE is 0 and KE is 11,760. Except to take off and hit the pool in the center she needs acceleration. Which is change in v/t. Neither of which I have. Where do I go from here?
  14. Nov 11, 2008 #13
    Here is what I thought if her Vi=19.79 m/s (not sure of units)
    then to solve for b, KE=0.5*60*19.79= 593.7. Right?
  15. Nov 11, 2008 #14
    a) h=1/2 g*t^2 is correct. Don't forget the 1/2.
    Solve for t and then use the equation for x (no acceleration on x, right) to find the initial velocity.

    b) Write conservation of energy (see post by JWDavid) . You don't need to calculate the kinetic energies. Just the difference.
    Hint: It will not depend on the initial velocity.
  16. Nov 11, 2008 #15
    h(20)= 1/2*9.8m/s*t^2
    Is time 15.1 s?
  17. Nov 12, 2008 #16
    No. t = sqrt(20*2/9.8).
    One thing I've found in the few months I've been doing physics, you have got to reality check you answers (ALMOST every time). 20m =~ 60 feet it doesn't take even a feather 15 seconds to fall 60 feet. The answer for time is gonna be somewhere around 2 seconds.
    Therefore your answer to a) is gonna be somewhere (again around) 3m/s, because it will be distance / time
  18. Nov 12, 2008 #17
    Your right, I just get so overwhelmed. The time is squared so it would be the square root of 15.1= 3.88s. I appreciate your help. I am just stuck, I don't know how to solve this equation. I just am not sure what information I need, and then what to do with it. I DO see that the height, mass and distance are important but I don't know how to "build" an equation to solve this.
  19. Nov 12, 2008 #18

    This is how i would approach the question

    Since you are dealing with two dimensional motion x,y break it up and look at them seperately

    First look at the y, component
    knowing gravity,and the intial vertical velocity you can work out the time taken to fall the distance to the ground using your basic kinematics formula
    y(final) = y(intial) + v_y(intial)*t - (1/2)*a_y*t2

    Now knowing the time to fall from the top to bottom you can work out the horizontal distance travelled in this time period. knowing the distance you can solve for "v_x(intial)" to reach the desired 6m.

    x(final) = x(intial) + v_x(intial)*t - (1/2)*a_x*t2

    I find it easier with these problems to break them up and look at the x and y components seperately, the key is knowing that the time taken to fall, is the time used to calculate the horizontal distance travelled while falling.

    You can use energy but it gets abit messier, in my experiance hope it heeps cheers
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