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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Sagot :

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Answer:

a. 3.73 m/s b. 27.8 m/s²

Explanation:

(a) Calculate his velocity (in m/s) when he leaves the floor.

Using the conservation of energy principles,

Potential energy gained by basketball player = kinetic energy loss of basket ball player

So, ΔU + ΔK = 0

ΔU = -ΔK

mg(h' - h) = -1/2m(v'² - v²)

g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor

Substituting the values of the variables into the equation, we have

9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)

9.8 m/s²(0.710 m) = -1/2(-v²)

6.958 m²/s² = v²/2

v² = 2 × 6.958 m²/s²

v² = 13.916 m²/s²

v = √(13.916 m²/s²)

v = 3.73 m/s

(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m

So, making, a subject of the formula, we have

a = (v² - u²)/2s

Substituting the values of the variables into the equation, we have

a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)

a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)

a = 13.913 m²/s²/(0.50 m)

a = 27.83 m/s²

a ≅ 27.8 m/s²

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