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Distance Question

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. (y = 0.56 m, and x = 5.8 m.) Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

    3-45alt.gif

    2. Relevant equations

    Velocity, Acceleration, Time, and Distance. We are using all of those, but in all truth, I have NO idea of what formula to use...

    3. The attempt at a solution

    I don't even know how to start the problem, it seems to me like some information is missing. The problem itself is sort of confusing, too much irrelevant information. Please please help me.
     
  2. jcsd
  3. Oct 2, 2011 #2

    lewando

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    What information seems irrelevant?
     
  4. Oct 2, 2011 #3
    The whole paragraph, which confuses me. The only information that it gives is the x and y values.
     
  5. Oct 2, 2011 #4

    lewando

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    The speed of the bullet is important, also the initial angle of the gun is important. All the information you need is there. You should focus first on the region inside the second building. Find out how long it took for the bullet to hit the ground after coming through the window. That is a good starting point.
     
  6. Oct 2, 2011 #5
    Alright, I used d=1/2*a*t^2. Put .56m for d and found t to be .34 s. How does that help me? Thanks for helping me by the way.
     
  7. Oct 2, 2011 #6
    I keep looking at the problem and I just can't figure out how using 1 section of the whole problem, the inside of the building, is going to help me find the rest of the problem. The only way I can think of is using ratios with the 340 m/s, but then again, that is constant. What formula am I able to use to find "something" outside of the building?
     
  8. Oct 2, 2011 #7

    gneill

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    For a projectile, which component of the velocity remains constant throughout the trajectory?

    For a projectile, what is the vertical velocity at the apex of the trajectory? So where's the apex located?

    How do you calculate the maximum vertical displacement of a projectile given the initial vertical component of the velocity?
     
  9. Oct 2, 2011 #8
    The horizontal velocity remains the same throughout the trajectory. I'm guessing that that is the key, but I don't know how to use velocity in graphs where distance is the x axis or where velocity is the slope of triangles, it throws me off.

    H is the apex of the bullet. It's one of the things I'm trying to find, but I can't, I would need the time it took to fall from the apex.

    I'm not sure of this one. To calculate H, the maximum vertical displacement, I need at least 2 variables, and I only have 1, gravity. Somehow I need to use that small triangle inside the building where the bullet hits, to find one of the bigger vectors, but I don't know how.
     
  10. Oct 2, 2011 #9

    gneill

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    Hint: You'll want to find the velocity components of the bullet when it struck the glass.

    You know that the horizontal velocity of the bullet is constant. So what's the time it took to go from the glass to the wall?

    When the bullet reached the glass it had some unknown vertical velocity. But in the time that the bullet traveled across the room to hit the wall it dropped distance y. What equation of motion is applicable for the vertical motion of a projectile with an initial vertical velocity, distance traveled, and time?
     
  11. Oct 2, 2011 #10
    Mmmm, the amount of time it took to touch the ground was .34s and the amount of time it took to hit the wall the moment it entered the building is .017s.

    And I don't know any formulas that only have the initial velocity and not the final in it. The basic d=v*t is the only one I know that gets the closest, but it's the change of velocity is v(f)-v(i).

    EDIT: Wait, wouldn't you need the original height, H, to find the time it took to touch the ground? The vertical velocity increased by the second with gravity. I used d=1/2*a*t^2, to find the time it took to touch the ground. But by the time the bullet reaches the window, it would of already be having some vertical velocity. I'm even more confused now...
     
    Last edited: Oct 2, 2011
  12. Oct 2, 2011 #11

    gneill

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    The problem doesn't say anything about the bullet hitting the floor. More likely it embedded itself in the wall. The bullet traveled from the glass to the wall at a constant horizontal velocity. What was that velocity? How long did it take to cover that distance?
     
  13. Oct 2, 2011 #12
    By the floor I mean where it stops.

    The constant horizontal velocity is 340 m/s. With t=d/v, d=.56m, and v 340 m/s. The amount of time it took to hit the wall is .00165s.
     
  14. Oct 2, 2011 #13

    lewando

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    Sorry, my bad. I meant the wall.
     
  15. Oct 2, 2011 #14

    gneill

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    But your d is the vertical height that the bullet traveled between the glass and wall, not the horizontal distance! Use the horizontal distance for horizontal motion.
     
  16. Oct 2, 2011 #15
    Argh, I had it right the first time, it's .017s. Now it makes more sense. But how would that help me?
     
  17. Oct 2, 2011 #16

    gneill

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    Well, now you know the time it took for the bullet to go from the glass to the wall. Now you can look at the vertical motion of the bullet in that same space.

    The bullet hit the glass with some (as yet unknown) vertical velocity component. Call it Vy. In the time it took to traverse the room it dropped y = 0.56m. What's an equation that describes the vertical distance traveled?
     
  18. Oct 2, 2011 #17
    d=v[itex]_{i}[/itex]t+.5at[itex]^{2}[/itex] ?

    Using that equation, I get that the v[itex]_{i}[/itex]=32.86 m/s. Am I on the right path?
     
    Last edited: Oct 2, 2011
  19. Oct 2, 2011 #18
    WOOOOO!!!! Finished the problem! Thank you guys so much! I really appreciate it.
     
  20. Oct 2, 2011 #19

    gneill

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    Excellent, another satisfied PF customer :smile:
     
  21. Oct 2, 2011 #20

    gneill

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    By the way, when you calculated the height of the gun, you did take into account that it's measured from the level of the bullet in the wall rather than the hole in the glass, right?
     
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