Question

a cannon ball is launched into the air with an upward velocity of 346 feet per second, from a 14 - foot tall cannon. the height h of the cannon ball after t seconds can be found using the equation ( h=-16t^{2}+346t + 14 ).
approximately how long will it take for the cannon ball to be 1112 feet high? round answers to the nearest tenth if necessary.
units:

how long long will it take to hit the ground?
units:

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Explanation:

Step1: Set up the equation for height \(h = 1112\)

Substitute \(h = 1112\) into \(h=-16t^{2}+346t + 14\).
We get \(1112=-16t^{2}+346t + 14\).
Rearrange to standard quadratic form \(16t^{2}-346t + 1098 = 0\).
Divide by 2: \(8t^{2}-173t + 549=0\).
Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 8\), \(b=-173\), \(c = 549\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-173)^{2}-4\times8\times549\).
\(\Delta = 29929-17568=12361\).
Then \(t=\frac{173\pm\sqrt{12361}}{16}\).
\(\sqrt{12361}\approx111.18\).
\(t_{1}=\frac{173 + 111.18}{16}=\frac{284.18}{16}\approx17.8\), \(t_{2}=\frac{173-111.18}{16}=\frac{61.82}{16}\approx3.9\).

Step2: Set up the equation for hitting the ground (\(h = 0\))

Substitute \(h = 0\) into \(h=-16t^{2}+346t + 14\).
We get \(16t^{2}-346t - 14=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 16\), \(b=-346\), \(c=-14\).
Calculate the discriminant \(\Delta=(-346)^{2}-4\times16\times(-14)\).
\(\Delta = 119716 + 896=120612\).
\(\sqrt{120612}\approx347.3\).
\(t=\frac{346\pm347.3}{32}\).
We take the positive root \(t=\frac{346 + 347.3}{32}=\frac{693.3}{32}\approx21.7\) (since \(t>0\) and \(\frac{346-347.3}{32}<0\) is rejected).

Answer:

For height \(h = 1112\) feet: \(t\approx3.9\) seconds and \(t\approx17.8\) seconds.
For hitting the ground (\(h = 0\)): \(t\approx21.7\) seconds.