Asked by Dan

A golfer hits a shot to an elevated green. The ball leaves the club with an initial speed of 16 m/s at an angle of 58' above the horizontal. If the speed of the ball just before it lands is 12 m/s, what is the elevation of the green above the point where the ball is struck?
Answered by Scott
initial K.E ... 1/2 m v^2 = 128 m

final K.E. ... 72 m

m g h = 128 m - 72 m

g h = 56 ... h = 56 / 9.8
Answered by Dan
I'm sorry, but where did 128m and 72m come from? Is that part of a formula?
Answered by Dan
As well as 56?
Answered by Dan
Never mind of the 56, sorry.
Answered by Dan
I've figured out the problem, thanks for your help.
Answered by Anonymous
but v vertical = v sin 58
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