A ball is thrown downward from a window 16.5 m from the ground with a velocity of 3.6 m/s. What is the velocity of the ball the moment it hits the ground? Round your answer to 2 decimal places.

To find the velocity of the ball the moment it hits the ground, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity (what we want to find)
u = initial velocity (3.6 m/s, thrown downward)
a = acceleration due to gravity (9.8 m/s^2, pointing downward)
s = distance traveled (16.5 m, downward direction)

Plugging in these values into the equation, we have:

v^2 = (3.6 m/s)^2 + 2(9.8 m/s^2)(-16.5 m)

v^2 = 12.96 m^2/s^2 + (-323.4 m^2/s^2)

v^2 = -310.44 m^2/s^2

To solve for v, we take the square root of both sides:

v = √(-310.44 m^2/s^2)

However, since we're looking for a velocity, we can ignore the negative sign because the velocity is always positive when the ball hits the ground. So,

v = √310.44 m^2/s^2

v ≈ 17.61 m/s

Therefore, the velocity of the ball the moment it hits the ground is approximately 17.61 m/s.