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A rock is thrown with height, direction, and inital velocity provided

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data

    A rock is thrown off a 100 m cliff at an angle of 38 degrees below horizontal with an initial velocity of 20.0 m/s.
    a) What is the total time in flight?
    b) What is the horizontal range?
    c) What is the max height reached by the rock?


    2. Relevant equations

    x = x_0 + v_x t
    y = y_0 + v_0y t - 1/2 g t^2


    3. The attempt at a solution

    The answers are
    a) t = 3.43 s
    b) R= 54.2 m
    c) 100m, trick question.


    I don't get how the answers are made (I struggle with motion in two directions), but I've had them provided to me in class. I'm confused as to the process to get the answers. Above are the equations the instructor used in class, but I'm also confused as to why he used them. My book states only three kinematic equations, and those are:
    v_f = v_0 + a t
    Δx = v_0 t + 1/2 a t^2
    v_f^2 = v_0^2 + 2 a Δx
    and their vertical identicals.

    Is it possible to get the answers using those three kinematic equations instead of the ones he used? Or can someone talk me through WHEN those two equations he used should be used? Like are they for projectile motion in two directions only or something?

    Thanks in advance!
     
  2. jcsd
  3. Feb 20, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The second equation from your book (highlighted) can be used to get both of the equations used by your instructor. (In one case the acceleration will be zero.)
     
  4. Feb 20, 2014 #3
    Ahhhh, I see now. I always thought the first equation was given by v_0 = {x_f - x_0}over{t}.

    So when I do it:
    I plug in my values into Δx = v_0 t + 1/2 a t^2, I get time to equal either 69.70 or -167.27 after using the quadratic equation. This is not right.
     
  5. Feb 20, 2014 #4
    Am I suppose to split the initial velocity into x and y components?
     
  6. Feb 20, 2014 #5
    I found it, had to split the initial velocity and use the horizontal component into my initial velocity when I did the quadratic.

    How am I suppose to know when to use the x or the y componants when I solve a problem like this?

    Also, since I found time (rather simple, too) how do I solve part B?
     
  7. Feb 20, 2014 #6
    BTW - when I use the range equation for part B, I get 39.6 m.
     
  8. Feb 20, 2014 #7
    NVM, found it. Had to plug the x-component of velocity into the first equation my instructor gave with the time we found.

    Thanks, this can be closed.
     
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