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How much time after his friend lets go of the apple does Jim need to fire the gun?

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Jim's friend climbs up a 25m tower and lets an apple fall straight down. Jim is standing 10 m away from tower and will shoot the apple just as teh apple is at a height of 1.5m, the same heightfrom which the gun was fired. The bullet has an initial speed of 40 m/s. How much time after friend lets go of apple does Jim need to fire the gun?

    2. Relevant equations


    3. The attempt at a solution

    I calculated the time it took for the bullet to get from inital position to position where it's supposed to meet the apple-40m/s divided by 10m since apple and gun are at the same height...I think that's right...so, I got 4 seconds.

    But I'm stuck on how to calculate the time it took for the apple to go from 20m to 1.5 m. The distance the apple will have to go would obviously be 18.5m and because it's falling straight down, wouldn't that mean a=-9.8 m/s^2?

    Thanks in advance!

    I may be posting like 3 more questions! :/
     
  2. jcsd
  3. Oct 5, 2007 #2
    First: 40 divided by 10 is .25, not 4

    I don't want to answer your question the way it's worded. As I'm interpretting it, the apple is supposed to be hit by the bullet when it's 1.5 meters above the ground. If you aim the gun horizontally, it will indeed traverse that horizontal distance in .25 seconds. However, the bullet will no longer be 1.5 meters above the ground as it's subjected to the same acceleration due to gravity as the apple.

    Thus, you're going to have to aim the gun at an angle. There are only two possible angles to shoot the gun at such that it'll be 1.5 meters above the ground 10 meters away. (i.e. range problems: ignoring air resistance, a cannon shooting a cannonball has maximum range at 45 degrees. Any shorter range can be achieved by one angle greater than 45 degrees, and another angle less than 45 degrees. Unfortunately, when you aim the gun at some other angle, you're going to decrease the horizontal component of the motion of the bullet, causing time to increase to beyond .25 seconds for it to arrive. Thus, the first thing you'll have to find is an angle such that your bullet will actually be in the right place.


    Does your teacher typically give problems this "tricky?" Yours is solvable, but a little more difficult than I've seen most teachers give their students.

    edit: I worked out the solution, and in reality, the effect of needing to aim slightly above the horizontal is almost negligible given the distance and the height. The total time for the bullet to arrive is still very close to .25 seconds. Continue to work from this point but you can probably note the flaw. If you shoot the bullet perfectly horizontal, you should be able to calculate the distance it falls in the y-direction due to gravity. (It'll miss about 30cm too low)
     
    Last edited: Oct 5, 2007
  4. Oct 5, 2007 #3
    sorry 1/4 is what I meant, so that part is correct?


    "Thus, the first thing you'll have to find is an angle such that your bullet will actually be in the right place."
    -YIKES! So, how do I do that?

    "...you should be able to calculate the distance it falls in the y-direction due to gravity. (It'll miss about 30cm too low)."

    That's my problem, I don't know how to calculate the distance it falls in the y-direction...


    This is so confusing, I apologize for my "dodoness." I really need a tutor...ASAP.

    And, this course is online, so I'm pretty much self taught because the lessons are vague.
     
  5. Oct 5, 2007 #4

    DaveC426913

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    drpizza, I know what you're getting at - the bullet falls too - but I suspect that you're overcomplicating it. I suspect that, for the sake of this question, that effect is to be ignored The gun can be fired horizontal and it is treated as if the bullet's path is straight and horizontal.

    I could be wrong - that's an awfully slow projectile.
     
    Last edited: Oct 5, 2007
  6. Oct 5, 2007 #5

    DaveC426913

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    Um. Care to check that again? :surprised


    (I think what you meant to say is that "the correct calculation is 10 divided by 40"... :tongue2:)
     
  7. Oct 5, 2007 #6
    "(I think what you meant to say is that "the correct calculation is 10 divided by 40"... )"

    Oh man, I've re-read my post 100 times, yes, TEN DIVIDED BY 40 is what I mean! I'm so flustered.....
     
  8. Oct 5, 2007 #7
    This is just a guess but...

    Since you are ignoring air resistance and other such anomalies and it doesn't explicitly state what angle the gun needs to be fired and hit the apple, I would ignore that....

    We know the gun needs to be fired in time for the apple to get hit at exactly 1.5 meters, we know the bullet takes .25 seconds to arrive once shot.

    What you need to do is calculate how far the bullet falls in .25 seconds, add that distance to 1.5 meters and then calculate how long it takes for the apple to reach that point. :)
     
  9. Oct 5, 2007 #8

    DaveC426913

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    Actually, it was drpizza who said that, and he was the one trying to tutor you...
     
  10. Oct 5, 2007 #9

    DaveC426913

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    The falling aspect drops out of your calculations.

    You know your bullet will take .25 seconds to reach the target.

    The bullet will fall exactly as fast as the apple, meeting it at the right point.

    Simply put: You know where to shoot. Aim directly at the apple when it is .25seconds from it's target point of 1.5 metres. Bullet and apple will fall the same distance in the intervening .25s.
     
  11. Oct 5, 2007 #10
    yeah sorry, didn't think it through enough....so it's just the time it takes to reach 1.5 meters, minus .25 seconds, if you aim directly at the apple....that's kind of a badly worded question.....
     
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