Posted by Wes on Saturday, February 2, 2013 at 6:36pm.
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.
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