If a pro basketball player has a vertical leap of about 20 inches, what is his hang time?

hang time = 0.6455 seconds.

Step-by-step explanation:

First of all you have to define hang time. If we don't agree on that, any answer is impossible. I think it means from the beginning of a jump that the player makes where he leaves the floor and lands on the floor again.

Briefly leaving the floor to landing on the floor again.

First you must convert 20 inches to feet

1 foot = 12 inches x feet = 20 inches

1/x = 12/20 Cross multiply

12*x = 1 * 20 Combine factors.

12x = 20 Divide by 12

12x/12 = 20/12 Do the division

x = 1.67 Feet

Solve for time.

Givens

vf = 0 a = - 32 feet/sec^2 d = 1.667 feet.

Formula

d = vf*t - 1/2 a * t^2

Solution

1.667 = 0*t - 1/2 (-32 feet/sec^2) * t^2

1.667 = 16 t^2 Divide by 16

1.667 / 16 = t^2 Do the division

0.1041667 sec^2 = t^2 Take the square root of both sides Switch

t^2 = 0.1041667

sqrt (t^2) = sqrt (0.1041667)

t = 0.3227 seconds.

Comment

That gets you to the peak of the jump. Now you must come down. You need to double the time to get the answer.

t = 2*0.3227

t = 0.6455 seconds.