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Projectile Motion

  1. Mar 15, 2005 #1
    This is my first year at studying physics and I understand th econcepts however I am having trouble applying these concepts to "problem questions". This is a question I urgently need help with:

    Lice, Bob, Charlie & Daniel are standing on the edge of a canyon which is 100m deep & 20m wide. On the opposite side of the canyon, there is a 10m high cave, 15m below the top of the cliff. They want to see who can throw a rock horizontally & get it into the cave. They each throw the rock with the following initial velocities: Alice 5m/s, Bob 10m/s, Cgarlie 15m/s, Daniel 20m/s. Whos rock makes it into the cave?

    I know I need to find the time it takes for a rock to be thrown n hit the top of the cave, and for the bottom. Then calculate the time it takes each person the throw there rock and hit the opposing wall. Whoevers is inbetween the top of cave and bottom of cave times, is the person who will get there rock into the cave. I also know I must somewhere use the displacement rule...I have no idea how i should begin, what rules are needed, and how to complete such a difficult question! Help would be greatly appreciated :smile:
     
  2. jcsd
  3. Mar 15, 2005 #2

    learningphysics

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    The trick is to deal with vertical and horizontal motion separately. First I'd draw a sketch, if a picture isn't already given.

    What is the height difference between where the kids are standing and the top of the cave? Once you know this you should be able to get the time to hit the top.

    What is the height difference between where the kids are standing and the bottom of the cave? Once you know this you should be able to get the time to hit the bottom.

    You can calculate the times for each person and find the times that are in between as you mentioned.

    However, instead of calculating the times for each person... you can instead use the time to hit the top and bottom, and use that time to calculate the velocity required to hit the top or bottom. And then any velocity in between will get through. I think this is a little quicker to do.
     
  4. Mar 15, 2005 #3

    tony873004

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    Is it to be assumed that the top of the cliff on the cave side is the same height as the top of the cliff that the rockthrowers are standing on?

    10 feet high, 15 feet below the top of the cliff? Is the top of the cave 15 feet below the top of the cliff, or is the bottom of the cave 15 feet below the top of the cliff, or is the center of the cave 15 feet below the top of the cliff?

    Let's assume the middle.

    [tex]y_f = y_i + v_{iy} + \frac {1}{2} g t^2[/tex]

    [tex] y_i=0[/tex] if you consider the top of the cliff to be 0. Forget about the 100 m deep part. If you consider the ground to be 0 and the top of the cliffs to be 100, you're just adding math that doesn't need to be there to the problem.
    [tex]v_{yi}=0[/tex] if they're throwing horizontal

    so your formula simplifies to

    [tex]y_f = \frac{1}{2} gt^2[/tex], where [tex]y_f[/tex] is the final y position below the top of the cliff.

    Let's use Alice's number of 5 m/s.

    [tex]y_f = \frac{1}{2}(9.8m/s^2) t^2[/tex]

    We need to know t to get any further. T is easy since the all of the initial velocity is in the x direction, and gravity can't touch the x direction. So [tex]x_i[/tex] is equal to the velocity in the x direction for the entire problem. And in the x direction, the other cliff is 20 m away.

    [tex]v = \frac{d}{t}[/tex] This should be intuitive. Just think of your car's speedometer. Your velocity might be 60 mi/h, where mi is a distance, and h is a time.

    Therefore, [tex]t=\frac{d}{v}[/tex]

    [tex]t=\frac{20 m}{5 m/s}[/tex] The meters cancel, and you're left with

    [tex]t=4 s[/tex]

    Now that you have t, you can plug it into the above formula:
    [tex]y_f = \frac{1}{2}(9.8m/s^2) t^2[/tex]

    [tex]y_f = \frac{1}{2}(9.8 m/s^2) *(4s)^2[/tex]

    the [tex]s^{2} [/tex] 's cancel

    [tex]y_f=78.4 m[/tex]

    Alice's rock will drop 78.4 meters by the time it reaches the other side. It will obviously miss the cave. Get an arm, Alice! Move over, it's Bob's turn...
     
  5. Mar 15, 2005 #4
    please dont answer questions fully. Youre taking away form the learning experience. Helpfull hints are encouraged but DO NOT solve the full problem.

    Regards,

    Nenad
     
  6. Mar 15, 2005 #5

    tony873004

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    Sorry, but I only solved a quarter of the problem. There's 3 more people. And I justified every step to help him understand, rather than just doing it like I would do it on a homework. It was my intention to make sure that he understood it, because if you don't understand these kinds of problems, physics gets real tough!

    He still needs to apply what I have explained to the 3 other people to get a complete answer worthy of turning in. And he also still has to resolve the height of the cave issue.
     
  7. Mar 15, 2005 #6
    he can look at what you have done and then just plug in the new numbers to solve the other three people. That is not understanding the problem, that is plug and play.

    Regards,

    Nenad
     
  8. Mar 15, 2005 #7

    tony873004

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    You're right. I should have left a little more for him to solve. :smile: I just hope the explanations help him understand these type of projectile problems a little better.
     
  9. Mar 15, 2005 #8
    I know some problems are very tempting to solve, and sometimes it is hard to resist solving a nice question or proof, but I have gotten in trouble for this kind of stuff before and now I see why.

    Regards,

    Nenad
     
  10. Mar 19, 2005 #9
    THANX FOR EVERY1s HELP!

    Thank-you everyone who atteptem to help me with my projectile Motion question. Hapilly though I was able to nut it out myself, which felt really great. Then I had all ur replys to let me know that indeed I had taken the correct path, so thanks

    Felicia
     
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