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Projectile Motion bag of mail

  1. Feb 8, 2009 #1
    I've been trying to do this question for ages, but I just can't seem to figure it out. I cant find the height of the second building.


    A bag of mail is catapulted from the top of a building 200 m above the ground with a velocity of 20 m/s at an angle of 15 degrees above the horizontal. If the mail is to land on the roof of another building 100 m away, how tall is the second building?
  2. jcsd
  3. Feb 8, 2009 #2
    Re: Projectile Motion, find max height, TWO POSSIBLE ANSWERS?

    See, you know the equation of the path of the projectile. You also know the horizontal distance (x-coordinate) of the top of the building wrt the point of launch. Substitute in the projectile eqation to get the reqd Y co-ordinate.
  4. Feb 8, 2009 #3
    Re: Projectile Motion, find max height, TWO POSSIBLE ANSWERS?

    Yea, I tried doing that, but i got the wrong answer for some reason. The height I got for the building was 135 m. I don't know what I'm doing wrong.
  5. Feb 8, 2009 #4
    Re: Projectile Motion, find max height, TWO POSSIBLE ANSWERS?

    Make sure of what you are taking as the origin. If the point of launch is the origin, then you will have to take into consideration the height of the first building.
  6. Feb 8, 2009 #5
    Re: Projectile Motion, find max height, TWO POSSIBLE ANSWERS?

    But how will in incorporate that into finding how tall the second building is. I was thinking i could use the final velocity of the x and y components to figure out the height. But I guess that would be wrong too
  7. Feb 8, 2009 #6
    This should make things clearer. How do you think the given information helps??
    Last edited: Feb 8, 2009
  8. Feb 8, 2009 #7
    I can't open the attachment. Sorry I'm new here so i really don't know things work.
    It says the attachment is pending an approval. Ill just wait.
  9. Feb 8, 2009 #8
    I'm going to be honest. I kinda had that same diagram drawn out. I still can't draw out an equation to find out how tall the second building is.
  10. Feb 9, 2009 #9
    Okay, you have that picture.

    Say that the origin of your coordinate system is at (0,0), so that the initial height is at (0,200) and the final is (100,H).

    [tex]\vec{r}=\vec{r}_{0} + \vec{v}_{0}t + \frac{\vec{a}t^{2}}{2}[/tex]

    I suspect you've seen this equation before. Split it into the y- and x-coordinates, and try to use the coordinates I wrote out earlier.
  11. Feb 9, 2009 #10
    This scenario lacks sufficient information for everyone to derive the same answer.

    Unlike the ground, buildings have edges, so rooftop dimensions are required if you intend to land a projectile on a target rooftop. However, no dimensions for the target rooftop were given, which opens Pandora’s Box to many variations of target building rooftop heights that can still be deemed correct.

    Via presumption, the target rooftop could possess dimensions that are nearly pinpoint in area just 100 meters from the launch point of the mailbag or the target rooftop could be as large as a city block, in which case, the greater the horizontal rooftop dimension, the greater can be the variation in height of the target building’s rooftop and still land the mailbag on its rooftop.

    Additionally, a specific point of impact upon the target building’s rooftop "within" the specified dimensions should be provided to ensure everyone derives the same answer.
  12. Feb 9, 2009 #11


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    Homework Helper

    It's catapulted at 20m/s at an angle of 15°

    So you know your horizontal velocity from θ. Dividing that into the horizontal distance of 100m gives the minimum flight time and from that you can determine the minimum height differential between the buildings. (You also know your vertical Vy from θ as well.)

    ΔH = Vy*t - 1/2*g*t2
  13. Feb 9, 2009 #12
    I tried LowlyPion's method and i got the answer correct. It was only a matter of using the same time for the vertical equation to get the answer. Thanks a lot. I really appreciate everyone's help
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