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Simple 2D ballistics

  1. Dec 4, 2009 #1
    Well, finals are coming around. I'm reviewing stuff I thought I learned at the beginning of the semester only to find out that I don't remember what I'm doing. It's pretty easy until I forget everything.
    I think I have a handle on this, but I just want some verification to make sure I'm doing it right.

    1. The problem statement, all variables and given/known data
    A man stands on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 60.0 m/s at an angle of 33.0° above the horizontal. Calculate a) the maximum height above the roof reached by the rock; b) the magnitude of the velocity of the rock just before it strikes the ground; c) the horizontal distance from the base of the building to the point where the rock strikes the ground.


    2. Relevant equations
    I'm a big fan of using these:
    Vf=Vo+a(t)
    (delta)x=Vo(t)+.5(a)(t^2)
    (delta)x=[Vf^2 - Vo^2]/2a

    and some basic trig



    3. The attempt at a solution

    For (a), I chopped up my initial vector into x and y components.
    Vox = 50.3202 m/s
    Voy = 32.6783 m/s

    Then, I figured I'd not only see how far up it would go, but how long it would take to get there.
    For time, I used:
    0=32.6783{m/s}-9.81{m/s^s}*t
    I wound up with 3.3113 seconds

    Next, I got the same result of 54.4278 meters (above the roof) whether I used this equation
    (delta)x=Vo(t)+.5(a)(t^2) <--using the time I had previously solved for
    or this one
    (delta)x=[Vf^2 - Vo^2]/2a <--ignoring time altogether

    Then, for (b) I added the 54.4 meters to my initial 30 and solved for a new time (this one for it's downward movement) with this one again:
    (delta)x=Vo(t)+.5(a)(t^2)
    84.4...=0+(.5*9.81*t^2)
    Netted myself another 4.14881 seconds

    Using this new time, I solved for my final y velocity with
    Vf=Vo+a(t)
    Vf=0+9.81*4.14881
    I wound up with 40.6998 m/s
    I verified by using this formula:
    Vf^2=Vo^2 + 2a(delta x)
    or rather
    Vf^2=0+(2*9.81*84.4...)


    Then, I used the handy formula for building one vector out of 2:
    [(x^2)+(y^2)]^(1/2) <-- somebody is gonna' have to teach me how to type a "square root" symbol
    [(50.3202^2 m/s) + (40.6998^2 m/s)]^(1/2)
    I now have a vector magnitude of about 64.7 m/s

    Lastly, for (c), I added my two times together
    4.14881+3.3113
    Very nearly 7.5 seconds in flight.

    I multiplied my initial x vector by the flight time to see how far it flew before it landed.
    50.3202 m/s * 7.47993 s

    So it looks like this guy threw a rock 376.4 meters.

    That's over three and a half football fields.
    Either this guy has one heck of an arm (60 meters a second initial velocity? Madness!), or I royally jacked this up.

    Am I crazy? More importantly, are my calculations correct?
    Something just seems out of place and I can't put my finger on it.
     
  2. jcsd
  3. Dec 4, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Your answer is correct.
     
  4. Dec 4, 2009 #3
    Yup, the man does have one heck of an arm.

    I am a senior high school student and I hate it when there are problems that do not even try to be semi-realistic. It leads me to believe that my answer is incorrect somehow.

    Heck the man in this problem broke the record pitching speed for a baseball.
     
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