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A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?


A. 1.13 s; 69.75 ft


B.1.13 s; 29.25 ft


C.2.25 s; 9 ft


D.1.13 s; 31.5 ft

Answer :

watermelonwonto
b is ur answer hope i helped

Answer:  B) 1.13 s; 29.25 ft

Step-by-step explanation:

Height of the ball h in feet after t seconds is given by the function

[tex]h=-16t^{2}+36t+9[/tex]

Now at maximum height [tex]\frac{\partial h}{\partial t}=0

=>\frac{\partial( -16t^{2}+36t+9)}{\partial t}=0[/tex]

=>[tex]\therefore -32t+36=0=>t=1.125\, seconds[/tex]

After rounding to nearest hundredth,[tex]t\simeq 1.13\, seconds[/tex]

Height , [tex]h_{max}=-16t^{2}+36t+9[/tex]

[tex]=>h_{max}=[-16(1.125)^{2}+36\times 1.125+9] feet=29.25 feet[/tex]

Therefore , t=1.13 s and  [tex]h_{max}=29.25 feet[/tex]

Thus correct option is B) 1.13 s; 29.25 ft

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