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 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.
To determine the time the player spends in the top 10.7 cm of his jump, we need to use the principles of motion and the equation of motion:
s = ut + 1/2at^2
where:
s = displacement
u = initial velocity
a = acceleration
t = time
In this case, the displacement is 10.7 cm, the initial velocity is 0 (as the player is momentarily at rest at the top of the jump), and the acceleration is due to gravity (-9.8 m/s^2).
First, we need to convert the displacement and initial velocity to meters, as acceleration due to gravity is usually given in m/s^2.
10.7 cm = 0.107 m
0 m/s = 0 m/s
Now we can rearrange the equation to solve for time:
s = ut + 1/2at^2
0.107 = 0t + 1/2(-9.8)t^2
0.107 = -4.9t^2
Next, we can solve for time by isolating t:
t^2 = 0.107 / -4.9
t^2 = -0.0218
The equation has a negative value, which indicates no real solution in this case. It means that the player does not spend any time within the top 10.7 cm of his jump.
Therefore, the time spent in the top 10.7 cm of the jump is effectively zero.