## Problem:

If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

## Solution:

Given: a = -9.8 m/s2 vf = 0 m/s (final velocity) d = 1.29 m (distance travelled) | Find:t (hang time) vi (initial velocity) |

**Formula: vf^2 – vi^2 = 2ad**

(0 m/s)2 = vi^2 + 2*(-9.8 m/s2)*(1.29 m)

0 m2/s2 = vi^2 - 25.28 m2/s2

25.28 m2/s2 = vi^2

vi = 5.03 m/s

0 m2/s2 = vi^2 - 25.28 m2/s2

25.28 m2/s2 = vi^2

vi = 5.03 m/s

To find hang time, find the time to the peak and then double it.

**Formula: vf = vi + a*t**

0 m/s = 5.03 m/s + (-9.8 m/s2)*t

-5.03 m/s = (-9.8 m/s2)*t

(-5.03 m/s)/(-9.8 m/s2) = t

t = 0.513 s

hang time = 1.03 s

-5.03 m/s = (-9.8 m/s2)*t

(-5.03 m/s)/(-9.8 m/s2) = t

t = 0.513 s

hang time = 1.03 s