Question

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.15 m above the floor in an attempt to get the ball

Answer 1

To calculate the velocity at which the basketball player must leave the ground to rise 1.15 m above the floor, we can apply the equations of motion for vertical motion. The player needs to leave the ground with a velocity of approximately 4.26 m/s.

Step-by-step explanation:

To calculate the velocity at which the basketball player must leave the ground to rise 1.15 m above the floor, we can use the equations of motion for vertical motion. Let's assume that the initial velocity of the basketball player is 0 m/s and the acceleration due to gravity is -9.8 m/s^2 (assuming upward as positive and downward as negative).

Using the equation: vf^2 = vi^2 + 2aΔy, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and Δy is the change in height, we can rearrange the equation to solve for vf:

vf = sqrt(2aΔy)

Plugging in the values, we get:

vf = sqrt(2*(-9.8)*1.15) m/s

vf ≈ 4.26 m/s (rounded to two decimal places)

Answer 2

the basketball player must leave the ground with a velocity of 4.748 m/s

Step-by-step explanation:

Given that data in the question;

From the third equation of motion;

v² - u² = 2as

such that;

u² = v² - 2as

where u is the initial velocity, v is the final velocity, s is the displacement and a is acceleration

so initial velocity of the basket ball player will be;

u = √( v² - 2as )

so from the question; s is 1.15 m and a = - 9.8 m/s² { player is under negative acceleration to get to the ball } and final velocity of the player will be 0.

so we substitute

u = √( (0)² - (2 × -9.8 × 1.15 )

u = √ -( - 22.54 )

u = √ ( 22.54 )

u = 4.748 m/s

Therefore, the basketball player must leave the ground with a velocity of 4.748 m/s