Final answer:
The expression for acceleration during the leg-straightening of a basketball player is found by calculating the player's velocity when leaving the floor, the acceleration while straightening the legs, and the force exerted on the floor. The velocity is approximately 4.20m/s, the acceleration is approximately 14.0m/s^2, and the force exerted on the floor is approximately 1540N.
Step-by-step explanation:
The expression for acceleration during the leg-straightening of a basketball player can be determined using the given information. The player lowers his body by 0.300m and then accelerates through this distance. To find the expression for acceleration, we need to calculate the player's velocity when he leaves the floor and the distance he goes from zero to that velocity.
Part (a):
To calculate the player's velocity when he leaves the floor, we can use the equation for velocity. The initial velocity is zero and the displacement is 0.900m. Using the equation v^2 = u^2 + 2as, we can rearrange it to find v:
v^2 = 0^2 + 2 * (-9.8m/s^2) * (-0.900m)
v^2 = 17.64m^2/s^2
v = √17.64m^2/s^2
v ≈ 4.20m/s
Part (b):
To calculate the acceleration while the player is straightening his legs, we can use the equation a = (v - u) / s. The initial velocity is zero, the final velocity is 4.20m/s, and the distance is 0.300m. Substituting these values into the equation, we get:
a = (4.20m/s - 0) / 0.300m
a ≈ 14.0m/s^2
Part (c):
To calculate the force exerted by the player on the floor, we can use Newton's second law of motion, F = ma. The mass of the player is 110kg and the acceleration found in part (b) is 14.0m/s^2. Substituting these values into the equation, we get:
F = 110kg * 14.0m/s^2
F ≈ 1540N