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

Sagot :

Answer:

1.03s

Given, acceleration a=−g=−9.8m/s

2

(minus sign for motion against gravity); final velocity (top most point) is v=0m/s

Vertical distance traveled d=1.29m

If u be the initial velocity, using formula v

2

−u

2

=2ad

0

2

−u

2

=2(−9.8)(1.29)

⟹ u=5.03m/s

If t be the time to peak.

Using v=u+at

0=5.03−(9.8)t

We get t=5.03/9.8=0.513s

So, the hang time will be double of peak time i.e. t

hang

=2×0.513=1.03s