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The equation \(y = -4.9t^2 + 23.8t\) describes the height (in meters) of a projectile launched from the ground at 23.8 meters per second. In how many seconds will the projectile first reach 28 meters in height?

 Jul 23, 2019
 #1
avatar+794 
+5

the answer would be 2.5 or if you would like the exact answer 2.429

 

edit

sorry did not read question it would be 2 seconds at 28 meters

 Jul 23, 2019
edited by travisio  Jul 23, 2019
edited by travisio  Jul 23, 2019
 #2
avatar+36915 
-1

The height is given by 'y'    set y = to 28

28 = -4.9t^2 + 23.8t     now solve for 't'      

-4.9t^2 + 23.8t - 28 = 0      Use the quadratic formula  (a = -4.9   b = 23.8    c = -28 ) to solve for 't'

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)    

 

  Note:  You will find TWO values of 't'..... the FIRST one is the projectile going UP....the second is the projectile coming DOWN...

 Jul 23, 2019

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