Final answer:
Using kinematic equations, the initial velocity required for a basketball player to rise 1.25 m above the floor is approximately 4.95 m/s, which is not listed in the provided options.
Step-by-step explanation:
To determine the initial velocity required for a basketball player to leave the ground and rise 1.25 m in an attempt to get the ball during the tip-off, we can use the kinematic equations that describe projectile motion under the influence of gravity. The motion of the player in the vertical direction can be analyzed separately from the horizontal motion. In this case, since we're only interested in the vertical component, we can use the formula that links the initial velocity (v), the height (h), and acceleration due to gravity (g), which is roughly 9.8 m/s2:
h = v2 / (2g)
Plugging in the height of 1.25 m and solving for v gives us:
v2 = 2 * 9.8 m/s2 * 1.25 m
v2 = 24.5 m2/s2
v = √(24.5 m2/s2)
v = 4.95 m/s
The closest option indicating the required initial vertical velocity to achieve a 1.25 m vertical leap by a basketball player is not listed among the provided choices (a) 1.25 m/s, (b) 7.80 m/s, (c) 11.0 m/s, or (d) 14.5 m/s. The actual value needed is approximately 4.95 m/s.