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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 ms2 downwards.

A
u = 10 ms1, t = 0.5 s
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B
u = 5 ms1, t = 1 s
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C
u = 15 ms1, t = 1.5 s
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D
u = 10 ms1, t = 1 s
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Solution

The correct option is B u = 5 ms1, t = 1 s
Given, distance travelled one side s = 1.25 m
and acceleration, a=10 ms2 (his motion is against gravity)
Final velocity, v=0 ms1 ( since he is at rest after takeoff)
Let 'u' be the initial velocity and 't' be the time taken.
From the third equation of motion, v2=u2+2as,
0=u22×10×1.25
u=5 ms1
From the first equation of motion, v = u + at,
0=510×t
t=510=0.5 s
Since the time taken for going up and coming down will be equal. Total time for which he remained in air is 1 second.

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