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# A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck.?

The ball leaves the club at a speed of 14.9 m/s at an angle of 38.5° above the horizontal. It rises to its maximum height and then falls down to the green. It rises to its maximum height and then falls down to the green. Find the speed.

Please help Bill or the Old Science Guy

### 1 Answer

- NCSLv 74 months agoFavorite Answer
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 x-value 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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