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The height h in feet of an object shot straight up with initial velocity v in feet per second is given by h = −16t^2 + vt + c, where c is the initial height of the object above the ground. A model rocket is shot vertically up from a height of 6 feet above the ground with an initial velocity of 22 feet per second. Will it reach a height of 10 feet? Identify the correct explanation for your answer.

 Feb 18, 2021
 #1
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v =  22      c  = 6

 

So   we   have  that

 

-16t^2  +  22t   +  6  = 10       

 

-16 t^2  + 22t  - 4  =  0           multiply through by  - 1

 

16t^2  - 22t   +  4   =  0        divide through   by 2

 

8t^2  - 11t  + 2 =  0

 

Using the quadratic formula

 

 

t  =    11  ± sqrt ( 11^2  - 4 * 8 * 2)                11  ± sqrt (57)

       ________________________   =       _____________

                   2 * 8                                               16

 

It  first reaches  10  ft      at     [  11  - sqrt (57)  ]  / 16   ≈  .216 sec

 

 

cool cool cool

 Feb 18, 2021

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