Abasketball 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.

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 

Hope this helps!
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It is d, all the other ones say incorrect answer .

What formula  this one  (x_/