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Billy shoots an arrow from 10 feet above the ground. The height of this arrow can be expressed by the equation $h=10-23t-10t^2$, where $t$ is time in seconds. If the center of a target is raised 5 feet off the ground, in how many seconds must the arrow reach the target in order for Billy to hit the bulls eye?

 Feb 12, 2019
 #1
avatar+17269 
+2

Substitute h = 5 and solve the equation for t

 

5 = 10-23t-10t^2

0 = 5 -23t-10t^2    Use quadratic formula to find t = .2 sec

 Feb 12, 2019
edited by ElectricPavlov  Feb 12, 2019
 #2
avatar+98090 
+1

0 = 5 - 23t - 10t^2      rearrange as

 

10t^2 + 23t - 5 =  0

 

Factor as

 

(5t - 1) ( 2t + 5)  = 0

 

Setting both factors to 0  and solving for t, we get

 

t = 1/5 sec = 0.2 sec       or   t = -5/2  sec  =  -2.5 sec

 

It's obvious that the first answer is correct

 

 

 

cool cool cool

 Feb 12, 2019
 #3
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0

Thanks! It worked. That problem didn't make much sense to me

 Feb 12, 2019

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