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Sagot :
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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