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. The equation that relates velocity, initial velocity, distance, and acceleration is:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
In this case, we know that the ball is thrown downward, so the acceleration due to gravity (a) will be -9.8 m/s^2 (negative because it acts in the opposite direction of the throw). The initial velocity (u) is given as 3.6 m/s and the distance (s) is the height from the window to the ground, which is 16.5 m.
Substituting these values into the equation:
v^2 = (3.6)^2 + 2(-9.8)(16.5)
Simplifying:
v^2 = 12.96 - 323.4
v^2 = -310.44
Since we are looking for the velocity, we ignore the negative sign:
v = √(310.44)
Calculating:
v ≈ 17.62 m/s
Therefore, the velocity of the ball the moment it hits the ground is approximately 17.62 m/s.