If Michael Jordan has a vertical leap of 1.29 m, then what is his take off speed and his hang time (total time to move upwards to the peak and then return to the ground)?
Answers
Answer:
Let the initial velocity be ‘u’.
Acceleration due to gravity, g = -9.8 m/s2 [downward]
Let the time of flight be ‘t’.
So, using
v2 = u2 + 2as
=> 0 = u2 + 2(-9.8)(1.29)
=> u = 5.03 m/s
Now,
s = ut + ½ at2
=> 0 = 5.03t – 0.5 × 9.8 × t2
=> t = 1.03 s
Explanation:
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Answer:
- His take off speed is 5.07 m/s (approx)
- His hang time is 1.014 second .
Explanation:
According to the Question
It is given that ,
- Michael Jordan has a vertical leap of 1.29 m
we have to calculate his take off speed & his hang time .
As we know that ,
Acceleration due to gravity is 10m/s² .
Since, the he reached at vertical peak so his final velocity is zero .
Calculating his take of speed
By using Kinematic Equation
- v² = u² + 2gh
putting the value we get
⇢ 0² = u² + 2×(-10) × 1.29
⇢ 0 = u² + (-20) × 1.29
⇢ -u² = -25.8
⇢ u² = 25.8
⇢ u = √25.8
⇢ u = 5.07 m/s (approx)
- Hence, his take off speed is 5.07m/s
Now ,
Calculating his hang time
Again by using Kinematics Equation
- v = u + gt
by putting the value we get
⇢ 0 = 5.07 + (-10) ×t
⇢ -5.06 = -10t
⇢ t = 5.07/10
⇢ t = 0.507 s
Since, it is given that total time to move upward to the peak and return to the ground will be
⇢ Total time = 2×0.507
⇢ Total time = 1.014 s
- Hence, the total time taken is 1.014 second.