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.

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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okpalawalter8 okpalawalter8 · Answer 2 · 2021-10-02T03:09:09+02:00

We have that the vertical height he will attain is mathematically given as

s=3.920m

From the question we are told

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

Generally the Newtons equation for the velocity is mathematically given as

V=at

Where

[tex]t_h=\frac{1.11}{2}[/tex]

t_h=0.555

Therefore

U=9.8*0.555s

U=5.43m/s

Generally the Newtons equation for the is mathematically given as

[tex]s=ut+1/2at^2[/tex]

Where

[tex]s=5.43-1/2(9.8)(0.555)^2[/tex]

s=3.920m

Therefore

The vertical height he will attain is mathematically given as

s=3.920m

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