Final answer:
The ball will be traveling at approximately 3.57 m/s when it hits the floor.
Step-by-step explanation:
To solve this problem, we can use the principle of conservation of energy. At the point where the ball is released, it has gravitational potential energy and no kinetic energy. As it falls and then reaches the floor, it loses potential energy and gains an equal amount of kinetic energy.
Using the equation for gravitational potential energy, PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the floor:
PE = KE
mgh = 1/2mv²
Since the mass cancels out, we can simplify the equation to:
gh = 1/2v²
Given that the initial velocity of the ball is 4.5 m/s and the initial height is 1.3 m, we can plug in these values to find the final velocity:
9.8 * 1.3 = 1/2v²
12.74 = v²
v ≈ 3.57 m/s
Therefore, the ball will be traveling at approximately 3.57 m/s when it hits the floor.