Patrick L. answered • 10/05/15

Tutor

5.0
(84)
A patient and effective Math and Computer/Electronics Tutor

set x=0 at the top of the cliff and x increases as the ball goes down.

1. The falling ball hits the ground at

25 = (1/2)*10* t^2 , note: (s = 0.5gt^2, g = 10)

t = sqrt(5), this implies the velocity is

v=gt = 10*sqrt(5) note: (v0=0)

2. the two balls cross when:

(1/2)*g*t^2 = s = 25 - vt + (1/2)*g*t^2, where v =10*sqrt(5) as indicated in 1 above

==> 25 = 10*sqrt(5)*t1

t1 = (1/2)*sqrt(5)

3. the falling ball would have traveled:

d = (1/2)*10*t1^2 = (1/2) *10*(1/4)*5 = 25/4 meters

I am in Cupertino. Contact me if you have questions.