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The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2 + 14t - 0.4 at time t (in seconds). As an improper fraction, for how long is the cannonball above a height of 4 meters?

 Jul 10, 2022
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-4.9t^2  + 14t - .04  =  4

 

-4.9t^2  + 14t  -.04 - 4  = 0

 

-4.9t^2 + 14t - 4.4  = 0           multiply through by -10

 

49t^2  - 140t + 44  = 0

 

Solving this with the quadratic formula  we get

 

t ≈ .36 sec

and

t ≈ 2.5 sec

 

It is above 4 m  for  about      2.5 - .36  =  2.14 seconds =  214/100 =   107/50 sec

 

 

cool cool cool

 Jul 10, 2022

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