Asked by chelsea
Use the function V(T) = 18T2. A basketball player has a vertical leap of 38.5 inches.
What is their hang time to the nearest hundredth of a second?
Hint: T is your time in seconds.
What is their hang time to the nearest hundredth of a second?
Hint: T is your time in seconds.
Answers
Answered by
drwls
If V = 18 T^2 ,
with V (the vertical height) in inches and T in seconds, then
T = sqrt(V/18)
If V = 38.5 inches, then
T = sqrt(38.5/18) = 1.463 seconds
"sqrt" means "square root of"
with V (the vertical height) in inches and T in seconds, then
T = sqrt(V/18)
If V = 38.5 inches, then
T = sqrt(38.5/18) = 1.463 seconds
"sqrt" means "square root of"
Answered by
chelsea
Thank you so much for helping me.I appreciate it
Answered by
Reiny
I must admit that I am caught totally unaware of the recent change in the law of gravity.
I think drwls did not see the inches unit.
I thought it was
height = 16t^2 , where height is in feet
so 38.5 = 38.5/12 ft = 3.208 ft
3.208 = 16t^2
t = .2005
t = √.2005 = .448 seconds to go up
so hang time = 2(.448) or appr 0.9 seconds
(Michael Jordan was timed at a 1 second hangtime, I looked it up)
I think drwls did not see the inches unit.
I thought it was
height = 16t^2 , where height is in feet
so 38.5 = 38.5/12 ft = 3.208 ft
3.208 = 16t^2
t = .2005
t = √.2005 = .448 seconds to go up
so hang time = 2(.448) or appr 0.9 seconds
(Michael Jordan was timed at a 1 second hangtime, I looked it up)
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