Answer:
(a) 4.15 m/s
(b) 34.5 m/s²
(c) 4.07 × 10³ N
Note: These answers have been rounded to three significant figures.
Explanation:
To solve the problem of the basketball player jumping straight up, we need to calculate the velocity when he leaves the floor, his acceleration while straightening his legs, and the force he exerts on the floor.
[tex]\hrulefill[/tex]
(a) Calculate the velocity when he leaves the floor[tex]\hrulefill[/tex]
Knowing the player reaches a height of h = 0.880 m above the floor. We can use the fourth kinematic equation below for vertical motion:
[tex]\boxed{\begin{array}{ccc}\text{\underline{The 4 Kinematic Equations:}}\\\\1. \ \vec v_f=\vec v_0+\vec at\\\\2. \ \Delta \vec x=\frac{1}{2}(\vec v_f+\vec v_0)t\\\\3. \ \Delta \vec x=\vec v_0t+\frac{1}{2}\vec at^2\\\\ 4. \ \vec v_f^2=\vec v_0^2+2\vec a \Delta \vec x \end{array}}[/tex]
[tex]\Longrightarrow \vec v_{f_y}^2=\vec v_{0_y}^2+2\vec a_y \Delta \vec y[/tex]
[tex]\Longrightarrow (0 \text{ m/s})^2=\vec v_{0_y}^2+2(-9.8 \text{ m/s}^2)(0.880 \text{ m})[/tex]
[tex]\Longrightarrow \vec v_{0_y}^2=-2(-9.8 \text{ m/s}^2)(0.880 \text{ m})[/tex]
[tex]\Longrightarrow \vec v_{0_y}^2=17.248 \text{ m$^2$/s$^2$}[/tex]
[tex]\Longrightarrow \vec v_{0_y}=\sqrt{17.248 \text{ m$^2$/s$^2$}}[/tex]
[tex]\therefore \vec v_{0_y}\approx \boxed{4.15 \text{ m/s}}[/tex]
Thus, the player has a speed of approximately 4.15 m/s leaving the ground.
[tex]\hrulefill[/tex]
(b) Calculate his acceleration while straightening his legs[tex]\hrulefill[/tex]
We can find the player's acceleration by considering the distance over which he accelerates from rest to his takeoff velocity. We will use the same kinematic equation from above, where:
- v_f = 4.15 m/s
- v_0 = 0 m/s
- Δy = 0.250 m
[tex]\Longrightarrow \vec v_{f_y}^2=\vec v_{0_y}^2+2\vec a_y \Delta \vec y[/tex]
[tex]\Longrightarrow (4.15 \text{ m/s})^2=(0 \text{ m/s})^2+2\vec a_y (0.250 \text{ m})[/tex]
[tex]\Longrightarrow 17.248 \text{ m$^2$/s$^2$}=(0.500 \text{ m})\vec a_y[/tex]
[tex]\Longrightarrow \vec a_y=\dfrac{17.248 \text{ m$^2$/s$^2$}}{0.500 \text{ m}}[/tex]
[tex]\therefore \vec a_y \approx \boxed{34.5 \text{ m/s}^2}[/tex]
Thus, the acceleration while he is straightening his legs is approximately 34.5 m/s².
[tex]\hrulefill[/tex]
(c) Calculate the force he exerts on the floor[tex]\hrulefill[/tex]
Using the acceleration found in part (b), we apply Newton's second law to find the force exerted by the player.
Given:
[tex]\Longrightarrow F = ma[/tex]
[tex]\Longrightarrow F = (118 \text{ kg})(34.5 \text{ m/s}^2)\\\\\\\\\Longrightarrow F = 4070.528 \text{ N}\\\\\\\\\therefore F \approx \boxed{4.07 \times 10^3 \text{ N}}[/tex]
Thus, the force he exerts on the floor is approximately 4.07 × 10³ N.
