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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.
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hmm i think the formula is not right. it think it is \[h(t)=-16t^2+v_0t+h_0\]
i am also going to assume that the ball is thrown upward from the ground, otherwise we cannot now the answer.
yes
sorry i am in a rush this is due in less than 20 minutes
ok well you need to know what value of \(-\frac{v_0}{32}\) will give you 196
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still not finding it
give me a minute, we can get it
try 112
\[h(t)=-16t^2+112t\] vertex at \(\frac{-b}{2a}=\frac{112}{32}=\frac{7}{2}\) and \(h(\frac{7}{2})=196\)
thanks
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yw
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