Question 214935
What is the hang time of a basketball player with a 35 inch reach?


Step 1.  The equation is given as {{{h=g*t^2/2}}} where h is the height, g is the gravity given as 32 ft/sec^2 and t is the time to travel a given height.


Step 2.  The hang time is 2*t (t to go up and anther t to go down)


Step 3.   Solve for t in Step 1 as follows:


{{{h=g*t^2/2}}}


{{{2h/g=t^2}}} where we multiplied by 2/g to both sides of the equation


{{{t=sqrt(2h/g)}}} where we took the square root to both sides of the equation and selected the positive time.


{{{t=sqrt(2*35/(32*12))}}}  where 12 inch = 1 feet or 35 inches is 35/12 feet


{{{t=0.18229}}}


{{{2t=0.3646}}}


Step 4.  Hang time is 0.3646 seconds.


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J