Posted by **Sarah** on Saturday, December 17, 2011 at 6:07pm.

A basketball player, standing near the basket to grab a rebound, jumps 65.6 cm vertically. How much time does the player spend in the top 10.7 cm of his jump

- Physics -
**drwls**, Saturday, December 17, 2011 at 6:22pm
We will treat the player as a point mass, and refer to where his center of mass is located. To raise the center of mass by H = 0.656 m, the TOTAL time spent in the air must be 2*sqrt(2H/g) = 0.731 s . Half of that time is spent going up and half coming down.

The time to fall 0.107 m from the max height of 0.656 m to 0.549 m is

t = sqrt(2*0.107)/g) = 0.1477 s

The total time spent above that altitude is twice that, or 0.295 s.

- Physics -
**Damon**, Saturday, December 17, 2011 at 6:26pm
top at .656 meters

get Vi, initial speed from conservation of energy

(1/2) m Vi^2 = m (9.81)(.656)

Vi^2 = 2 (9.81)(.656)

Vi = 3.59 m/s

Now do a new problem, how fast is he going at h = .549 meters?

h = .549

(1/2) m v^2 + mgh = (1/2) m Vi^2

v^2 = Vi^2 - 2 g h

v^2 = (3.59)^2 - 2*9.81*.549

v = 1.45 meters/second

That is our initial condition for the top part

How long to top from there?(0 velocity up)

0 = 1.45 - 9.81 t

t = .148 s

total up and down 2 times t = .297 s

- Physics -
**Steve**, Saturday, December 17, 2011 at 6:27pm
How long does he stay in the air?

Well, his jump-up time is the same as if he had fallen from .656m

s = 1/2 at^2

.656 = 4.9t^2

t = .3659 sec

so, he rose for .3659s and then fell for .3659s

His velocity at landing is the same as it was at takeoff:

v = at

= 9.9 * .3659

= 3.5858 m/s

So, the equation of motion giving his height is

h = 3.5858t - 4.9 t^2

Note that h=0 at t=0 and t=.7318

Now, how long was he above 65.6-10.7 = 54.9cm?

Translate the graph down .549 and solve

3.5857t - 4.9 t^2 - .549 = 0

He was above 54.9cm from t=.2181 to t=.5136, or a total of .2955 seconds.

ute problem.

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