Answer:
hang time = 0.6455 seconds.
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.