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Homework Help: Kicking rock of a building

  1. Jul 25, 2007 #1
    kicking rock of a building....

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

    A boy kicks a rock at a velocity of 19.8 m/s horizontally off the top of a 64 meter high building. How far from the top does the rock hit the flat ground? (neglecting air resistance)

    2. Relevant equations

    sorry, i don't know how to type some symbols

    v^2 = v(initial)^2 + 2a(change in x)
    a(of y)= -9.8
    v^2(for y) = v(initial of y)^2 + 2a(change in y)

    3. The attempt at a solution

    not even sure how to begin. Is 19.8 m/s the initial velocity? how do I relate the equations? Do I need to use sine or cosine of the angle?

    The answer is 71.6 meters, for the record.
  2. jcsd
  3. Jul 25, 2007 #2


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    Science Advisor
    Homework Helper
    Gold Member

    19.8m/s is the initial velocity and is given in the horizonatl direction. Ther is no y component of the initial velocity. You've got to use your motion equations in each direction, and note thatt in the x direction, there is no acceleration, since no forces act in that direction during the flight.
  4. Jul 26, 2007 #3

    1) initial position and initial velocity
    This is a 2-D problem so you have x (horizontal) and y (vertical) components

    initial position : [tex]x_i = 0 e_x + 64 e_y[/tex]
    initial velocity : [tex]v_i = 19.8 e_x + 0 e_y[/tex]

    Gravity works along the y-axis, which you implemented correctly (-9.81)

    2) So applying Newton's second law gives us :

    [tex]a_i = 0 t^2/2 + 19.8 t[/tex]
    [tex]a_y = -9.81 t^2/2 + 0 t + 64[/tex]

    Do you understand these equations ?

    If so, solve them for t and you will have your answer

  5. Jul 26, 2007 #4
    I am still not getting it...

    I tried using the quadratic formula to solve for the time, using

    64 (y) = [19.8 (initial velocity) x t] - (1/2)(9.8)t^2

    I came up with an answer of 2.02 seconds.

    then I used:

    x = [v (initial velocity) x time] + (1/2) x (-9.8) x (time squared)

    came up with - 19.98.
    which is not the correct answer, nor can a change in x be negative
  6. Jul 26, 2007 #5

    Doc Al

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    Staff: Mentor

    You must take care to treat horizontal and vertical motion separately. What's the y-component of the initial velocity? (Recall that the rock is kicked horizontally.)

    Also, take care with the signs. Use + for up and - for down. The full equation for the vertical position is:
    [tex]y_f = y_i + v_{i,y} t - (1/2) (9.8) t^2[/tex]
  7. Jul 26, 2007 #6
    ok, i think i got it.... i was looking for how v for the x direction and v for the y direction were related..... thanks for your help guys : )
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