Question

Michael Jordan has a vertical leap of 1.25 m.
What is his takeoff speed and for how much time he will be in the air? Take a = 10 ms-2 downwards.
u = 10 ms-1, t = 1 s
u=5 ms-1,t = 15
u=15 ms-'. t = 1.5 s
u = 10 ms-!,t = 0.5 s

please ans this fast​

Answer 1

Answer:

Given, acceleration a=−g=−9.8m/s2 (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 v2−u2=2ad

02−u2=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. thang=2×0.513=1.03s 

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