SOLUTION: I am having a problem with a word problem for my math class. The problem is: If a basketball player has a vertical leap of about 20 inches, what is his hang time? Use the hang time
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Question 184077: I am having a problem with a word problem for my math class. The problem is: If a basketball player has a vertical leap of about 20 inches, what is his hang time? Use the hang time function V=48T^2. Here is what the example says to do.
To solve for T, first multiply by 1/48 to get T^2=20/48. Because hang time is positive, you can take the positive square root get T=squart20/48.
I get that part, what I don't get is how did they get that T=squart20/48=0.6 seconds. How would I get the seconds if the equation is T=squart35/48? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! If a basketball player has a vertical leap of about 20 inches, what is his hang time? Use the hang time function V=48T^2.
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Use basic algebra rules to solve for T, (isolate T by itself on the left)
write it:
48T^2 = V
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Substitute 20 for T
48T^2 = 20
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Divide both sides by 48
T^2 =
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Find the square root of both sides to change T^2 to just T
T =
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Enter into a calc, it will = .64549.. rounded to .6 sec
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You would do it exactly the same way if it were 35/48
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Does this make sense to you now?
