Question

10 multiple choice 2 points the formula ( h = vt - 16t^{2} ) gives a good approximation of the height in feet that an object will reach in t seconds, when it is projected upward with an initial speed of v feet per second. if a ball is thrown vertically upward with an initial velocity of 42 ft per second, at what times will it be 14 feet above the ground? (round your answers to the nearest hundredth of a second.) ( t = 0.21, t = 2.05 ) ( t = 0.39, t = 2.23 ) ( t = 0.60, t = 2.44 ) ( t = 0.53, t = 2.37 )

Explanation:

Step1: Substitute values into formula

Given \(v = 42\) and \(h = 14\), substitute into \(h=vt - 16t^{2}\). We get \(14 = 42t-16t^{2}\). Rearrange to \(16t^{2}-42t + 14=0\). Divide by 2: \(8t^{2}-21t + 7 = 0\).

Step2: Use quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 8\), \(b=-21\), \(c = 7\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-21)^{2}-4\times8\times7=441 - 224=217\).
Then \(t=\frac{21\pm\sqrt{217}}{16}\).
\(\sqrt{217}\approx14.73\).
\(t_{1}=\frac{21 + 14.73}{16}=\frac{35.73}{16}\approx2.23\).
\(t_{2}=\frac{21-14.73}{16}=\frac{6.27}{16}\approx0.39\).

Answer:

\(t = 0.39,t = 2.23\) (Option 2)