1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Freely Falling Bodies - Grrrr!

  1. Sep 17, 2005 #1
    Hey all -

    I have tried and tried to work this problem but I guess I am missing something (I can submit my answer several times to see if I am right and I have been wrong every time) -

    Here is the problem -

    A pellet gun is fired straight downward from the edge of a cliff that is 13 m above the ground. The pellet strikes the ground w/ a speed of 30 m/s . How far above the cliff edge would the pellet have gone had it been fired straight upward?

    My answer = 45.9 m

    I have attached the equation I used (to show super and subscripts) , and that equation is rearranged to solve for y (distance). My final velocity is 0 since at the peak of its height its velocity = 0 .

    Any help would be greatly appreciated!!

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Sep 17, 2005 #2
    You are on the right track. You may just be making a calculation error. You might try finding the time to reach a velocity of zero with v = v0 - at, and then use the distance formula. The numbers should agree.
     
  4. Sep 17, 2005 #3

    Fermat

    User Avatar
    Homework Helper

    The gun is fired downwards at a velocity of u, say, and strikes the ground with a velocity of v.

    Suppose the gun is fired vertically upwards at a velocity of u. It will reach a height = h, say. Then, when it returns to the cliff top again it will be travelling at the same speed, u, but in the opposite direction. And it will strike the ground with a velocity of v. Like in the first scenario.

    So, your problem resolves down to finding the height an object will fall from, h, if it reaches a final velocity of v when it hits the ground.
    Subtract the cliff top height from h to answer your question.
     
  5. Sep 17, 2005 #4
    Yes, I believe I went this route at one point - the time equals 3.06 seconds w/ that formula and plugged back into the distance formula I get 45.9 m (Formula is y= 1/2(v0 + v)t)

    Do you think I am still missing something here or should I send an e-mail to my instructor and ask about the problem? I doubt the answer listed there would be wrong and mine is right...
     
  6. Sep 17, 2005 #5
    Amazing... so I subtract the height of the cliff (13 m) from what I got for the height the pellet achieves above the cliff (45.9 m) and I get a correct answer of 32.9 m!

    So I guess does this mean the question asked at the end (How far above the cliff edge would the pellet have gone had the pellet been fired straight upward?) is wrong? Because it still seems like 45.9 m should be the height above the cliff, but I guess it can't be if 32.9 m is smaller and is the correct answer...

    Man, crazy...
     
  7. Sep 17, 2005 #6

    Fermat

    User Avatar
    Homework Helper

    I'm afraid not :frown:

    The gun velocity is 25.4 m/s.
    The pellet would have reached this velocity if it had fallen a distance of 32.9m.
    If the pellet had been fired upwards, from the top of the cliff, at 25.4 m/s, it would have reached a height of 32.9m, above the cliff top. Then it would have fallen back down again, reaching a velocity of 25.4 m/s when it reaches the cliff top again. It would then travel a further 13m downwards and hit the ground with a velocity of 30m/s.
     
  8. Sep 17, 2005 #7
    I see... well, I appreciate the help! Thanks much...

    -
    Morgan
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Freely Falling Bodies - Grrrr!
  1. Freely falling bodies (Replies: 1)

  2. Freely falling bodies (Replies: 3)

  3. Freely Falling Bodies (Replies: 2)

  4. Freely falling bodies (Replies: 3)

  5. Freely Falling Bodies (Replies: 2)

Loading...