Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 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 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.
Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 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 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.
Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 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 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.
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12. A stone is dropped from the top of a cliff. It is seen to hit the ground below
after 3.55 s. How high is the cliff?
13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming
no air resistance, what is the speed of the ball just before it strikes the ground?
14. Estimate (a) how long it took King Kong to fall straight down from the top
of the Empire State Building (280m high), and (b) his velocity just before
"landing".
Useful equations
For Constant Velocity:
V =>
D
X = V₁t + Xo
For Constant Acceleration:
Vr = V + at
X = Xo+Vot +
v=V+2a(X-Xo)
\prom = V +V
V velocity
t = time
D Distance
X = Final Position
Xo Initial Position
V = Final Velocity
Vo Initial Velocity
a = acceleration
For free fall
Yf
= Final Position
Yo Initial Position
g = 9.80
m
$2
For free fall:
V = V + gt
Y=Yo+Vo t +
+gt
V,² = V₁²+2g (Y-Yo)
V+Vo
Vprom=
2
6
Human Physiology: An Integrated Approach (8th Edition)
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