Question

A basketball player achieves a hang time of 1.11 s in dunking the ball. What vertical height will he attain? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.

Answer 1

since h.t is the total time the athlete is in the air, t/2 is the time taken for ascent or descent (t/2 + t/2 = t)

perhaps the initial jump vertical velocity u achieved is from the equation t/2 = u/g

therefore u = 9.8 (1.11/2) = 5.439 m/s

height = u²/2g = 5.439² / 19.6 = 1.5093225 m

Ans: 1.509 m correct to iv s.f


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