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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 ms−2 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 ms−1, t = 1 sGiven, distance travelled one side s = 1.25 m and acceleration, a=−10 ms−2 (his motion is against gravity) Final velocity, v=0 ms−1 ( 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=u2−2×10×1.25 ⇒u=5 ms−1 From the first equation of motion, v = u + at, 0=5−10×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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