Yahoo Answers: Answers and Comments for A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck.? [Physics]
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From Liam
enUS
Wed, 18 Sep 2019 15:19:00 +0000
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Yahoo Answers: Answers and Comments for A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck.? [Physics]
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https://answers.yahoo.com/question/index?qid=20190918151900AAEgUEh
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From NCS: Welcome to Yahoo!Answers.
I'm not Bill or...
https://answers.yahoo.com/question/index?qid=20190918151900AAEgUEh
https://answers.yahoo.com/question/index?qid=20190918151900AAEgUEh
Wed, 18 Sep 2019 18:41:14 +0000
Welcome to Yahoo!Answers.
I'm not Bill or the Old Science Guy, but I can help.
Find the FINAL speed? Well, we have the initial KE:
KE = ½mv² = ½ * M * (14.9m/s)² = M * 111m²/s²
When it reaches the green, it has 3 m worth of potential energy that it didn't have before (because the green is elevated):
GPE = mgh = M * 9.8m/s² * 3m = M * 29.4m²/s²
the final KE is
KE' = KE  GPE = M * 111m²/s²  M * 29.4m²/s² = M * 81.6m²/s²
and this
KE' = ½ * M * V²
where V is the final velocity
so
½ * M * V² = M * 81.6m²/s²
mass M cancels
V² = 163 m²/s²
V = 12.8 m/s ◄
OR
you could just use
v² = u² + 2as = (14.9m/s)² + 2 * 9.8m/s² * 3m = 163 m²/s²
as before.
I THOUGHT you were going to ask "How far away does the ball land?"
This can be done in one step using the trajectory equation:
y = h + x·tanΘ  g·x² / (2v²·cos²Θ)
where y = height at xvalue of interest = 3 m
and h = initial height = 0 m
and x = range of interest = ???
and Θ = launch angle = 38.5º
and v = launch velocity = 14.9 m/s
Dropping units for ease (x is in meters):
3 = 0 + x*tan38.5º  9.8x² / (2*14.9²*cos²38.5º)
0 = 3 + 0.7954x  0.0360x²
This quadratic has roots at
x = 4.83 m ← ball at 3 m height and rising
and x = 17.3 m ◄ ball at 3 m height and falling
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