1. Nov 16, 2009

songoku

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

Thanks

2. Nov 16, 2009

pynergee

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.

3. Nov 16, 2009

rl.bhat

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.

4. Nov 18, 2009

songoku

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 :

$$\frac{15}{2+x}=\frac{5+5t^2}{2}$$

$$x=\frac{6}{1+t^2}-2$$

$$\frac{dx}{dt}=\frac{-12t}{(1+t^2)^2}$$

For t = 1 :

$$\frac{dx}{dt}=-3~ ms^{-1}$$

Am I right?

Thanks