Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Physics 20-1 Final Review: Projectiles

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data
    This question is something we got in our review booklet and I HATE projectile motion.

    An object is thrown horizontally with a velocity of 18 m/s from the top of a cliff. If the object hits the ground 100 m from the base of of the cliff, how high is the cliff?

    2. Relevant equations

    d = v1t + 1/2at^2, v = d/t

    3. The attempt at a solution

    I try to break it up into components(vertical and horizontal) but nothing seems to work.

    I used the equation d = 1/2v1 + 1/2at^2 to solve for t, because obviously the initial velocity is 0 m/s, but i end up with a far-fetched answer.
  2. jcsd
  3. Jan 14, 2009 #2
    for some reason i have the same question, if you help me in my two threads, ill help you with yours. I'm beast at projectile motion
  4. Jan 14, 2009 #3
    so we're bargaining now? lol.
  5. Jan 14, 2009 #4
    1.You don't have any acceleration in the x direction so you can use that to find t.
    The fall ends when the object has traveld 100 m in the x direction at a constant velocity
    2. The equation is not correct d=v0t+(1/2)a*t2
  6. Jan 14, 2009 #5
    i still dont get it
  7. Jan 14, 2009 #6
    okay its thrown horizontally so no Vi in the Y direction
    g is -9.8
    you have distance in the x direction for 100m
    you need the velocity in the x direction and now you solve for time vx = 18ms
    T = dx/vx so 100/18 = T in seconds

    then now solve for distance in the y direction
    D = viy*t +1/2(-9.8)(t^2) viy = 0 so solve for T and then solve for D

    i think this should be right but im not sure

    now you gotta help me more im still confused :(
  8. Jan 14, 2009 #7
    You can use a frame of reference [tex](X,Y)[/tex] with origin [tex](0,0)[/tex] in the cliff and the point [tex]P(0,s_{oy})[/tex] is where the projectile will be thrown.
    Equation of motion respect to X and Y are:

    s_x=v_ox*t + s_{ox}\\
    s_y=\frac{1}{2} * g * t^2 + v_{oy}*t+s_{oy}

    but in according to my frame of reference [tex]s_{ox}=0[/tex]. Nevertheless [tex]v_{ox}=0[/tex] and [tex]v_{oy}=0[/tex]. You can make the equation of motion explicit in the following way (t disappeared):

    s_y=\frac{1}{2} * g * \frac{s_x^2}{v^2} + s_{oy}

    but where the projectile falls in ground



    [tex] - \frac{1}{2} * g * \frac{s_x^2}{v^2} = s_{oy} [/tex]

    if you replace [tex]s_x[/tex] and [tex]v[/tex] given by the problem you will have [tex]s_{oy}[/tex] which is the height of the cliff from which the stuff is thrown (if I have done no error it should be approximately 151.39 m).
    Last edited: Jan 14, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook