Question

If a pro basket ball player has a vertical leap of about 30 inches, what is the hang time? Use the hang time function v=48T to the second power

Answer 1

since the apex of his jump is 30 inches we can use this to backtrack

Answer 2

i got no idea what a hangtime function is tho

Answer 3

V=48T^2 ?

Answer 4

-16t^2 +v0 t = 30/12 right?

Answer 5

thats the position function well close to it

Answer 6

or rather the vertex of the parabola this makes is at 0,30

Answer 7

.. im at a loss for it at the moment

Answer 8

y = -16(t- n)^2 + 30/12

y = -16(t^2 -2t x +n^2) + 15/6

y = -16t^2 +32n t +(-16n^2 + 15/6), the ( ) part is the initial height
which is 0 right?


-16n^2 + 15/6 = 0

(-96n^2 + 15)/6 = 0

n^2 = 15/96 = 5/32

n = sqrt(5/32) if im right ... still aint determined if im right tho lol

Answer 9

y = -16t^2 +32(5/32) t

y = -16t^2 +5t = 0

t(-16t + 5) = 0 when t=0 and t=5/16 is my best guess

Answer 10

its wrong of course, but then i cant make sense of the questions as stated

Answer 11

a better thought might be to consider the time it takes to drop from 30 inches and double that

Answer 12

-16t^2 + 30/12 = 0

-16t^2 + 15/6 = 0

-16(6)t^2/6 + 15/6 = 0

(-96t^2+15)/6 = 0 when t^2 = 15/96

Answer 13

15/96 = 5/32

t = sqrt(5/32) , which should be half the time starting at the downward part of the jumpers arc thru the air, so double that to get the full time

2 sqrt(5/32)

Answer 14

.7906 seconds if we are counting in seconds :) and thats my best offer