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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 \(6\) meters?

 Feb 5, 2020
 #1
avatar+17 
+1

maybe just set this function equal to 6 and solve for t, the result will be two cases.

t1 nearly equal 0.57s

t2 nearly equal 2.29s

hope it can help :)

 Feb 5, 2020
edited by noobfromvn26  Feb 5, 2020
 #2
avatar+107011 
+2

6 = -4/9t^2 + 14t  - 0.4     rearrange as

 

4.9t^2 - 14t + 6.4  = 0       multiply through by 10

 

49t^2  - 140t + 64  =  0      factor

 

(7t  - 4) ( 7t - 16)  = 0

 

Set  each factor to 0  and solve for t  and we have that

 

t = 4/7  sec    and  t  = 16/7 sec

 

So.....the time it is above 6 m  is

 

(16/7) - (4/7)  = 

 

12/ 7    sec

 

 

cool cool cool

 Feb 5, 2020

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