Posted by Lauren on Friday, April 22, 2011 at 2:50pm.
a) KE at bottom= PE at top
1/2 m v^2=mgh solve for v.
where h is the distance from the starting point to the end point.
b) 1/2 m v^2=mg(h-2R)
c) at the top of the loop.
v^2/r=g as a minimum, so figure v^2.Then, using that v^2, 1/2 mv^2=mg(heighttostart) I dont understand the question.
e) figure v, that is the horizontal speed.
then figure the time to fall H
H= 1/2 g t^2 find t.
horizontal distance= v*t
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