Physics
posted by Sarah on .
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

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. 
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 
How long does he stay in the air?
Well, his jumpup 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 65.610.7 = 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.