Respuesta :
Answer:
[tex] 16t^2 +12 t -240 =0[/tex]
And we can use the quadratic formula given by:
[tex] t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where [tex] a = 16, b = 12, c =-240[/tex]. And replacing we got:
[tex] t = \frac{-12 \pm \sqrt{(12^2) -4*(16)(-240)}}{2*16}[/tex]
And after solve we got:
[tex] t_1 = 3.516 s , t_2 = -4.266s [/tex]
Since the time cannot be negative our final solution would be [tex] t = 3.516 s[/tex]
Step-by-step explanation:
We have the following function given:
[tex] h(t) = -16t^2 -12 t +240[/tex]
And we want to find how long after the ball is thrown will it strike the ground, so we want to find the value of t that makes the equation of h equal to 0
[tex] 0 = -16t^2 -12 t +240[/tex]
And we can rewrite the expression like this:
[tex] 16t^2 +12 t -240 =0[/tex]
And we can use the quadratic formula given by:
[tex] t = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]
Where [tex] a = 16, b = 12, c =-240[/tex]. And replacing we got:
[tex] t = \frac{-12 \pm \sqrt{(12^2) -4*(16)(-240)}}{2*16}[/tex]
And after solve we got:
[tex] t_1 = 3.516 s , t_2 = -4.266s [/tex]
Since the time cannot be negative our final solution would be [tex] t = 3.516 s[/tex]
