If michael jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time

The formula for the vertical height, h, attained with a vertical take off speed of u is
u² - 2gh = 0
where
g = 9.8 m/s², acceleration due to gravity.

Because h = 1.29 m, therefore
u² = 2*9.8*1.39 = 27.244
u = 5.2196 m/s

Also, the time of flight, t, required when the vertical take-off speed is u is given by
ut  - 0.5gt² = 0.
Therefore
5.2196t - 0.5*9.8*t² = 0
5.2196t - 4.9t² = 0
t(5.2196 - 4.9t) = 0
t = 0, or t = 5.2196/4.9 = 1.0652 s

t = 0 corresponds to take-off.
t = 1.0652 s corresponds to landing.
The hang time is 1.0652 s.

Take-off speed = 5.22 m/s (nearest hundredth)
Hang time  = 1.065 s (nearest thousandth)

The answer is 63.!

Where the white dots are is where your points should be plotted.a translation is simply just moving something from one place to another. in this case, you would move each plot point of the original shape 2 units up, and 5 units to the left.