Question

19 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 48 ft per second, at what times will it be 13 feet above the ground? (round your answers to the nearest hundredth of a second.)
( t = 0.51 )
( t = 2.91 )
( t = 0.23 )
( t = 2.63 )
( t = 0.44 )
( t = 2.84 )
( t = 0.12 )
( t = 2.52 )
( t = 0.30 )
( t = 2.70 )

Explanation:

Step1: Substitute values into formula

Given \(h = vt-16t^{2}\), \(v = 48\), \(h = 13\). So \(13=48t - 16t^{2}\). Rearrange to \(16t^{2}-48t + 13=0\).
For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 16\), \(b=-48\), \(c = 13\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate discriminant

\(\Delta=b^{2}-4ac=(-48)^{2}-4\times16\times13=2304 - 832=1472\).

Step3: Calculate \(t\) values

\(t=\frac{48\pm\sqrt{1472}}{32}\).
\(\sqrt{1472}\approx38.37\).
\(t_1=\frac{48 + 38.37}{32}=\frac{86.37}{32}\approx2.70\).
\(t_2=\frac{48-38.37}{32}=\frac{9.63}{32}\approx0.30\).

Answer:

\(t = 0.30\), \(t = 2.70\)