Physics
posted by Wes on .
Wes and Lindsay stand on the roof of a building. Wes leans over the edge and drops an apple. Lindsay waits, 1.25s afrer Wes releases his fruit and throws an orange straight down at 28 m/s. Both pieces of fruit hit the ground simultaneously. Calculate the common height from which the fruit were released. Ignore the effects of air resistence.

This is a duplicate post. Please don't clutter the website with identical posts

I would just do this by looking at Lindsay's orange, and use conservation of kinetic energy to solve this problem. KE=PE=1/2(mv^2)=mgh and solve for h.
g=9.8m/s^2
1/2(mv^2)=mgh
1/2(v^2)=gh
The masses are the same so they cancel out.
(1/2(28m/s)^2/9.8m/s^2)=h 
That method will give the wrong answer unless you use the kinetic energy change when it hits the ground.

I saw that after I did, I should have had him tackle it with the kinematic equations, but I believe you corrected this in a later post. Consequently, I didn't bother to post again saying ignore the setup.

Holy shzniit, this is so cool thank you.