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?

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

As well as 56?

Never mind of the 56, sorry.

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

but v vertical = v sin 58

To find the elevation of the green above the point where the ball is struck, we can use the principles of projectile motion.

Let's break down the information given:

Initial velocity (v₀) = 16 m/s
Launch angle (θ) = 58°
Final velocity (v) = 12 m/s

First, we need to calculate the horizontal and vertical components of the initial velocity (v₀x and v₀y).

v₀x = v₀ * cos(θ)
v₀x = 16 m/s * cos(58°)
v₀x = 8 m/s

v₀y = v₀ * sin(θ)
v₀y = 16 m/s * sin(58°)
v₀y = 13.82 m/s

Next, we can determine the time of flight (t) for the ball. Since the horizontal motion is unaffected by gravity, the time taken to reach the peak of the trajectory is equal to the time taken to fall from the peak to the ground.

The time taken to reach the peak (t₁) can be calculated using the vertical component of velocity:

v₀y = g * t₁
t₁ = v₀y / g
t₁ = 13.82 m/s / 9.8 m/s²
t₁ ≈ 1.41 s

The total time of flight (t) is twice the time taken to reach the peak:

t = 2 * t₁
t = 2 * 1.41 s
t ≈ 2.82 s

Now, we can find the vertical displacement (Δy) of the ball. The vertical displacement can be calculated using the formula:

Δy = v₀y * t - (1/2) * g * t²

Substituting the values:

Δy = 13.82 m/s * 2.82 s - (1/2) * 9.8 m/s² * (2.82 s)²
Δy ≈ 22.97 m

Therefore, the elevation of the green above the point where the ball is struck is approximately 22.97 meters.

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