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A ball is thrown into the air with an initial upward velocity of 42 ft/s. It height h in feet after t seconds is given by the function h (t)=-16t^2+42t+4.  What is the ball's maximum height?

 Mar 20, 2021
 #1
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We  have the  form

 

h(t)  =  at^2  + bt   + c

 

The  time for the  max  height is   -b/ ( 2a)  = = -42/ ( 2 * -16)  =  -42 / -32 = 21/16 sec

 

And  the  max  height  is  - 16 ( 21/16)^2  +  42(21/16)  + 4  =   31.5625 ft

 

 

cool cool cool

 Mar 20, 2021

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