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?

Answer 1

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

Answer 2

I'm sorry, but where did 128m and 72m come from? Is that part of a formula?

Answer 3

As well as 56?

Answer 4

Never mind of the 56, sorry.

Answer 5

I've figured out the problem, thanks for your help.

Answer 6

but v vertical = v sin 58