A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.8 m/s at an angle of 53.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

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h(t) = 3.00 + 17.8 sin53.0˚ - 4.9t^2

find t when h=0

Then, recall that the upward speed is

vy = 17.8 sin53.0˚ - 9.8t
vx = 17.8 cos53.0˚

the final v is of course found by

v^2 = vx^2 + vy^2

To find the speed of the ball just before it lands, we need to analyze the projectile motion of the ball. We can break down the motion into horizontal (x) and vertical (y) components.

First, let's look at the horizontal component of the motion. Since there is no horizontal force acting on the ball after it is hit, the horizontal velocity remains constant throughout the motion. We can find the horizontal velocity using the initial velocity and the launch angle.

Horizontal component (x):
Vx = V0 * cos(θ)

where:
Vx is the horizontal velocity,
V0 is the initial velocity (17.8 m/s), and
θ is the launch angle (53.0°).

Vx = 17.8 m/s * cos(53.0°)
Vx = 17.8 m/s * 0.6018
Vx ≈ 10.70 m/s

Now, let's focus on the vertical component of the motion. The initial vertical velocity can be found by multiplying the initial velocity by the sine of the launch angle.

Vertical component (y):
Vy = V0 * sin(θ)

Vy = 17.8 m/s * sin(53.0°)
Vy = 17.8 m/s * 0.7986
Vy ≈ 14.23 m/s

At the maximum height, the vertical component of the velocity becomes zero. We can determine the time it takes to reach the maximum height by dividing the initial vertical velocity by the acceleration due to gravity (g) in the negative direction.

0 = Vy - g * t_max

t_max = Vy / g
t_max = 14.23 m/s / 9.8 m/s^2
t_max ≈ 1.45 seconds

Since the ball reached the maximum height in the same amount of time it takes to descend to the ground, the total time of flight (T) is twice the time to reach the maximum height.

T = 2 * t_max
T ≈ 2 * 1.45 seconds
T ≈ 2.90 seconds

Now, we can determine the vertical component of the velocity just before the ball lands. During the descent, the vertical velocity increases due to the acceleration due to gravity.

Vy_final = -g * T
Vy_final = -9.8 m/s^2 * 2.90 seconds
Vy_final ≈ -28.42 m/s

Finally, we can calculate the total speed just before the ball lands by using the Pythagorean theorem to find the magnitude of the final velocity vector.

V_final = √(Vx^2 + Vy_final^2)
V_final = √(10.70 m/s)^2 + (-28.42 m/s)^2
V_final = √(114.49 m^2/s^2 + 809.12 m^2/s^2)
V_final = √(923.61 m^2/s^2)
V_final ≈ 30.4 m/s

Therefore, the speed of the ball just before it lands is approximately 30.4 m/s.