1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinematics about shadow

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    a basket ball is dropped from a height of 10 m and at a horizontal distance of 2 m from a light pole. A light source at the top of the light pole is 15 m above the ground. How fast is the shadow of the ball moving along the ground one second after the ball is dropped? it can be assumed that t seconds after the ball is released, the distance dropped is given by s = 5t2


    2. Relevant equations
    kinematics (I suppose)


    3. The attempt at a solution
    I don't have any ideas..

    This is the thing that I've done :

    after 1 s :
    s = 5 m
    v = u + at = 9.8 ms-1

    I don't know what to do about the shadow....

    Thanks
     
  2. jcsd
  3. Nov 16, 2009 #2
    Related rates problem? If so, you have to come up with a formula that relates the horizontal with the vertical of a triangle. Then differentiate that to come up with an equation that you can solve dx/dt for, which is what you are looking for.
     
  4. Nov 16, 2009 #3

    rl.bhat

    User Avatar
    Homework Helper

    Using horizontal distance and the height of then projection find the horizontal velocity v.
    After time t the horizontal displacement of the ball is v*t. Let the length of the shadow be x, and the vertical displacement ball is 5*t^2.
    After time t the vertical position of ball from the light is = .....?
    Draw the line joining the ball and the light and extend it to the ground. Now you can get two similar triangles. Then take the ratio of the proportionate sides. And find dx/dt.
     
  5. Nov 18, 2009 #4
    Hi pynergee and rl.bhat
    I think the horizontal velocity and displacement of the ball is zero because the ball is dropped so the trajectory of the ball is vertically downward, not parabolic?

    This is what I've tried. Assuming that the ball only has vertical movement, from similar triangles I got :

    [tex]\frac{15}{2+x}=\frac{5+5t^2}{2}[/tex]

    [tex]x=\frac{6}{1+t^2}-2[/tex]

    [tex]\frac{dx}{dt}=\frac{-12t}{(1+t^2)^2}[/tex]

    For t = 1 :

    [tex]\frac{dx}{dt}=-3~ ms^{-1}[/tex]

    Am I right?

    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinematics about shadow
  1. Kinematics about ramp (Replies: 3)

Loading...