Answer:
[tex] 8t^2 -48 t -7=0[/tex]
[tex] x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where:
a= 8, b =-48, c = -7 and replacing we got:
[tex] t_1 = \frac{48 - \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = -0.1425[/tex]
[tex] t_2 = \frac{48 + \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = 6.142[/tex]
And since the time cannot be negative the correct answer for this cae would be t = 6.142 s
Step-by-step explanation:
If the question is: When will the object hit the ground at when the time is how many seconds?
We have the function for the heigth given by:
[tex] h(t) = -16t^2 +96 t +14[/tex]
And we want to find the time where [tex] h(t) =0[/tex] so we can set like this the condition:
[tex] 0 =-16t^2 +96 t +14[/tex]
And we can rewrite the expression like this:
[tex] 16t^2 -96 t -14 =0[/tex]
We can divide both sides of the equation by 2 and we got:
[tex] 8t^2 -48 t -7=0[/tex]
And we can use the quadratic equation to solve for t like this:
[tex] x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where:
a= 8, b =-48, c = -7 and replacing we got:
[tex] t_1 = \frac{48 - \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = -0.1425[/tex]
[tex] t_2 = \frac{48 + \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = 6.142[/tex]
And since the time cannot be negative the correct answer for this cae would be t = 6.142 s
Step-by-step explanation:
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