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Mathematics 14 Online
OpenStudy (anonymous):

How fast would a ball have to be thrown upward to reach a maximum height of 196 ft? [Hint: Use the discriminant of the equation 16t2 − v0t + h = 0.

OpenStudy (anonymous):

hmm i think the formula is not right. it think it is \[h(t)=-16t^2+v_0t+h_0\]

OpenStudy (anonymous):

i am also going to assume that the ball is thrown upward from the ground, otherwise we cannot now the answer.

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

sorry i am in a rush this is due in less than 20 minutes

OpenStudy (anonymous):

ok well you need to know what value of \(-\frac{v_0}{32}\) will give you 196

OpenStudy (anonymous):

still not finding it

OpenStudy (anonymous):

give me a minute, we can get it

OpenStudy (anonymous):

try 112

OpenStudy (anonymous):

\[h(t)=-16t^2+112t\] vertex at \(\frac{-b}{2a}=\frac{112}{32}=\frac{7}{2}\) and \(h(\frac{7}{2})=196\)

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

yw

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