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Projectile motion:

  1. Sep 17, 2009 #1
    Hello hello, i have a question for you tonight:

    a ball on a 1.5m high table moving at a constant 5m/s rolls of the table.

    1. where does it land
    2. what are the Vx and Vy values JUST before it hits the ground
    thanks so much!

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    @pbdude: im sorry..i didnt really get what you said and i made a topic...thanks for helping though :)
     
  2. jcsd
  3. Sep 17, 2009 #2
    Hey there! I will ask you similar questions as the other thread to start you off:

    Since we know that it rolls off horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y ?

    Also, what are the relevant kinematic equations?
     
  4. Sep 17, 2009 #3

    okay, (Vo)y=0 because its only going forward.

    other formulas?

    Vx=Vcos(theta)
    Vy=Vsin(theta)
    Dx=Vx(t)
    Range/(Dy?)=V²(sin2(theta))/9.8
    Dy=Vy(t) + .5at²
     
  5. Sep 17, 2009 #4
    Okay, Don't worry about the 'range' formula. Let's stick to just:

    Dy=Vy(t) + .5at²
    Vf=Vo + at
    Vf2=Vo2 + 2a(df-do)

    Okay?

    We also know ay=g=-9.81 m/s2

    So, can we apply EQ 1 in the y-direction to find out how long the ball was in the air for?
     
    Last edited: Sep 17, 2009
  6. Sep 17, 2009 #5
    okay, i can do that but i dont know EQ 3 :(
    Dy=Vy(t) + .5at²
    1.5=0+4.9t² (my teacher uses 9.8 m/s²)

    sooo t=.5533?
     
  7. Sep 17, 2009 #6
    so, i hope double posting is allowed but
    the EQ Vf= Vo + at means that Vf=Vxf or Vyf, and Vo=Vox or Voy?
     
  8. Sep 17, 2009 #7
    Now how far did it land from the table (the x-direction)?

    You have Vox, and you have the amount of time it travels at that speed. You should be able to answer part (1) now.

    Yes. I gave you the 'general' form of the EQs. You can apply them to each direction independently as needed.
     
  9. Sep 17, 2009 #8
    OHHH

    right :P

    5m/s for Vox and .5533 for t

    sooo 2.766 M and the answer in the book is 2.8 OH YEAH


    thanks SO much :)
     
  10. Sep 17, 2009 #9
    so how do i find Vx and Vy? if i dont know theta?
     
  11. Sep 17, 2009 #10
    Great! For part (2) you know that Vox is constant, so V_fx is the same.

    To find V_fy, apply the kinematic EQ: V_fy= V_oy + at

    where a=g and 't' you have found.
     
  12. Sep 17, 2009 #11
    ahhh thanks again :D
     
  13. Sep 17, 2009 #12
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