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2 Projectile Motion Problem

  1. Sep 27, 2007 #1
    #1
    A bullet traveling 800 m/s horizontally hits a target 180 m away. How far does the bullet fall before it hits the target ?
    ---------------------------------------
    My work:
    X
    d = 180 m
    Vx = 180 m/s
    t = .225 s

    Y
    a = -9.8 m/s2
    t = .225 s (time in x direction = time in y direction)
    I'm stuck here ! Help please



    #2
    A baseball was hit at 45 m/s at an angle 45' above the horiztontal .
    a. HOw long did it remain in the air ?
    b. How far did it travel horizontally ?
    -------------------------------------

    My work:
    Vx = 31.8 m/s = Vy
    ay = -9.8 m /s2
    I only got to that point ! Help please
     
  2. jcsd
  3. Sep 27, 2007 #2

    Mentz114

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    Gold Member

    Here's a hint - a body being accelerated at a ms^-2 travels distance s = 0.5at^2 m in time t.

    [tex]s = \frac{1}{2}at^2[/tex]
     
  4. Sep 27, 2007 #3
    for problem #1 or #2?
     
  5. Sep 27, 2007 #4

    Mentz114

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    Gold Member

    In problem #1, you know how long the bullet was in the air ... which is enough to solve the problem. Do you know that a falling body accelerates at g = 32 m/sec/sec ?
     
  6. Sep 27, 2007 #5
    problem 1: vx=800 m/s not 180m/s
    time is correct.

    think about this:
    a=-9.8 m/s^2
    t=.225
    vo=0
    x=vot+.5at^2, as mentz114 stated: vo=0 so x=.5at^2
    you have a and t, now find x

    problem 2:
    vx=vy=31.8m/s
    ay=-9.8n/s^2
    ball is hit? can you visualize the arch that represents the ball in the air?
    ball has to travel from hit point to maximum height
    vf=vo+at, as maximum height, what is vf? once you have that, solve for time
    this time is the time from hit to max height, what do you need to do to find time from hit to max to horizontal?
    when you find the time from hit to max and back to horizontal, d=vt, solve
     
  7. Sep 27, 2007 #6
    Wow ... it look really easy ! Thanks a lot silvashadow, Mentz114
    I got them all !
     
  8. Oct 5, 2007 #7
    It's probably a typo, but g is 9.8m/s/s not 32 :)

    These make it look so easy! Gaaah
     
  9. Oct 5, 2007 #8
    You have it correct now just find the vertical distance... y=1/2gt*2.. with the time you found.


    So far you are correct... No solve how long it would take the ball to travel straight up at that velocity. t= vf-vi/g... that is to the maximum height... so you must double it for the time up and down.

    part b you now have the time it was in the air so now use... d=Vx X T... Make sure you use the total time you just solved for.
     
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