In the vertical jump, an Kobe Bryant starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat Kobe as a particle and let ymax be his maximum height above the floor. Note: this isn't the entire story since Kobe can twist and curl up in the air, but then we can no longer treat him as a particle.
To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.
NOTE: I got something like 70.7% of the time, but no matter how i write the answer it tells me that I'm wrong. I tried .700, 70.0, and .707 and none were good. PLEASE HELP =]
The answer was 4.82. I tried solving it, but I ran out of attempts inputting .707,etc. I still have no idea how
it was a format and i ran out of tries. Basically it wanted me to divide .707 by .293. The worst part is that it was worth a lot =[ o well
The time he takes to go up from y(max) /2 to the y{max } is the same as the time he falls from y [max ] to y [ max] /2.
Let y max = H and then y max = H/2
t1^2 = 2g H/2. ===========1
Simnilarly,
Time it takes him to go from the floor to that height
t2^2 = 2g H======================2
1divided by 2 gives
(t1/t2)^2 = 1/2 = 0.5
t1/t2 = 0.707 neaarly 70 %.
t1 = 70% t2.
Thus, for 70% of the total time of going up, he is in the upper half of the total height.
Hence he seems to hang in the air
It's .707/.293 because its the ymax/ (1-y max) or something like that.
So the reason it is .707/.293 is you're finding the ratio between how much time it was above h/2 vs. bellow h/2.
To calculate the ratio of the time Kobe Bryant is above ymax/2 moving up to the time it takes him to go from the floor to that height, we need to consider the motion of his vertical jump.
Let's break down the vertical jump into two phases: upward motion and downward motion. During the upward motion, Kobe accelerates upwards until he reaches his maximum height, ymax. During the downward motion, he accelerates downwards due to gravity until he reaches the floor.
In the absence of air resistance, we can use the kinematic equations of motion to determine the times taken for each phase. The key equation we will use is:
v = u + at
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
1) Upward motion:
During the upward motion, Kobe starts from rest (u = 0) and reaches maximum height (v = 0).
The acceleration is given by a = -g (where g is the acceleration due to gravity, approximately 9.8 m/s^2).
Using the equation v = u + at and rearranging for t:
0 = 0 + (-9.8)t
t = 0 s
So, the time taken for Kobe to reach his maximum height during the upward motion is 0 seconds.
2) Downward motion:
During the downward motion, Kobe starts from rest (u = 0) at ymax/2 and accelerates downwards due to gravity (a = -9.8 m/s^2).
Using the equation v = u + at and rearranging for t:
0 = 0 + (-9.8)t
t = 0 s
So, the time taken for Kobe to descend from ymax/2 to the floor during the downward motion is also 0 seconds.
Now, let's calculate the ratio of the time Kobe is above ymax/2 moving up to the time it takes him to go from the floor to that height:
time_ratio = (time_above_ymax/2_upward) / (time_floor_to_ymax/2_downward)
= (0 s) / (0 s) = undefined
From the calculations, it appears that the ratio of the time Kobe is above ymax/2 moving up to the time it takes him to go from the floor to that height is undefined. This means that Kobe does not spend any time at exactly ymax/2 during his vertical jump.
Please note that this result assumes no air resistance, and also simplifies the motion to treat Kobe as a particle. In reality, as mentioned in the question, Kobe can twist and curl up in the air, which would introduce additional complexities to the motion.