A golfer hits a shot to a green that is elevated 3.30 m above the point where the ball is struck. The ball leaves the club at a speed of 17.3 m/s at an angle of 49.0˚ 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.

See Related Questions: Sun, 10-21-13, 9:10 PM.

To find the speed of the ball just before it lands, we need to break down the motion of the ball into two parts: the vertical motion and the horizontal motion. Since there is no air resistance, the vertical motion of the ball can be analyzed separately from the horizontal motion.

Let's start by calculating the time it takes for the ball to reach its maximum height. We can use the vertical motion equation:

vf = vi + at

Where:
vf - final vertical velocity (0 m/s at maximum height)
vi - initial vertical velocity (17.3 m/s at launch)
a - acceleration due to gravity (-9.8 m/s^2)
t - time

Rearranging the equation to solve for time (t), we have:

t = (vf - vi) / a

Substituting the given values, we get:

t = (0 - 17.3) / (-9.8) = 1.766 seconds

Next, let's find the maximum height reached by the ball. We can use the vertical motion equation:

h = viy * t + 0.5 * a * t^2

Where:
h - maximum height
viy - initial vertical velocity (17.3 * sin(49.0˚))
t - time
a - acceleration due to gravity (-9.8 m/s^2)

Substituting the given values, we have:

h = (17.3 * sin(49.0˚)) * 1.766 + 0.5 * (-9.8) * (1.766)^2 = 9.343 meters

Now, let's find the time it takes for the ball to fall from the maximum height to the ground. We can use the equation:

h = 0.5 * a * t^2

Where:
h - maximum height (9.343 meters)
a - acceleration due to gravity (-9.8 m/s^2)
t - time

Solving for time (t), we get:

t = sqrt(2h / a)

Substituting the given values, we have:

t = sqrt(2 * 9.343 / (-9.8)) = 1.437 seconds

Now that we know the total time of flight, including the upward and downward motion, we can find the speed of the ball just before it lands.

We'll focus on the horizontal motion for this calculation. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion. The horizontal distance traveled can be calculated using the equation:

d = vix * t

Where:
d - horizontal distance traveled
vix - initial horizontal velocity (17.3 * cos(49.0˚))
t - total time of flight

Substituting the given values, we have:

vix = 17.3 * cos(49.0˚) = 11.130 m/s

Finally, using the equation for horizontal distance, we can find the speed of the ball just before it lands:

d = vix * t
vfx = d / t

Substituting the given values, we have:

vfx = (11.130 m/s) * (1.437 s + 1.766 s) = 32.418 m/s

Therefore, the speed of the ball just before it lands is approximately 32.42 m/s.