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Projectile angles and speed (x and Y) near impact

  1. Jul 31, 2012 #1
    Thought I was done but not the case.....

    1. The problem statement, all variables and given/known data
    Alright so as stated a couple days ago... "projectile is shot from the edge of a cliff 125 m above the ground @ an initial speed of 65 m/s at an angle of 37° with the horizontal from the cliff."
    ---- As helped earlier I have calculated the time it takes for the projectile to hit the ground (10.4 seconds) and on my own the distance it will travel (540 M). But there is more and I am stuck...

    " at the moment before impact with the ground find; 1)the horizontal and vertical components of it's velocity, 2)the magnitude of the velocity and 3) the angle made by the velocity vector with the horizontal."


    2. Relevant equations

    Not quite sure here

    3. The attempt at a solution

    I have figured out the Vxf by

    Vxo= (65 m/s)(cos 37°)= 51.9 m/s which I equate to the velocity in the x direction for the whole flight so it can apply at impact as well as at launch.

    But since the Vyf is in the downward direction I am lost as how to proceed from here to finish the rest of the problem.
     
  2. jcsd
  3. Jul 31, 2012 #2
    To find Vfy you need to use one of the kinematic equations for constant acceleration and apply it on the vertical component of velocity.
     
  4. Jul 31, 2012 #3
    So would I go something like this?

    Vyf = Vyo + at

    Vyf = (65 m/s)(sin 37) + (-9.8 m/s^2)(10.4s)

    = 39.1 m/s - 101.92 m/s

    = -62.82 m/s

    answer in the book is -63.1 m/s so I am comfortable with that.

    So the overall magnitude of the velocity I would go...

    √[(51.9)^2 + (-62.8)^2
    = 81.5 m/s

    for the overall angle on that beast would I then use...

    sin θ = 51.9/ -62.8
    = -55.7°

    Book says 50.6 below, which is understood because it is negative but I am off with something here.. what am I missing?
     
  5. Jul 31, 2012 #4
    Take a second look at this.
     
  6. Jul 31, 2012 #5
    Flip it???

    so go...

    tan θ = -62.8/51.9

    = -50.4°

    Think that will jive... thank you for the help Villyer!
     
  7. Jul 31, 2012 #6
    You're welcome! And it wasn't just flipping it, you also wrote sin the first time:b
     
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