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Mathematics 16 Online
OpenStudy (anonymous):

Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in feet, of the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t) and explain what the solution represents in the context of the problem? H(t)=-16t^2 + 80t + 96 g(t) = 31 + 32.2t

OpenStudy (anonymous):

I really need help

OpenStudy (anonymous):

i'm not sure about the table but if you want, you can solve it algebraically to find the solutions H(t) = g(t) \(-16t^2 + 80t + 96 = 31 + 32.2t \) \(-16t^2 + 80t - 32.2t + 96 - 31=0\) \(-16t^2 + 47.8t + 65=0\) then use the quadratic formula to solve for the values of t \(\Large t=\frac{ -b \pm \sqrt{b^2-4ac} }{2a}\) where a=-16, b=47.8, c=65

OpenStudy (anonymous):

I got -.83 and .14 is that right?

OpenStudy (anonymous):

Do H(t) and g(t) intersect when the projectile is going up or down and how do you know?

OpenStudy (anonymous):

they will intersect at two points, since there's a solution when you set H(t)=g(t) and for the the value of "t" i got a different answer t is approximately 4.002 and -1.015

OpenStudy (anonymous):

sorry late response, i'm away from my keyboard awhile ago

OpenStudy (anonymous):

so where will they intersect?

OpenStudy (anonymous):

when H(t)=G(t) which means that the values for "t" that you just found out

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