Solution:
Given that,
The height of a ball thrown into the air after t seconds have elapsed is:
[tex]h = -16t^2 + 40t + 6[/tex]
What is the first time, t, when the ball will reach a height of 20 feet?
Substitute h = 20
[tex]20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0[/tex]
Solve by quadractic formula
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}[/tex]
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=8,\:b=-20,\:c=7[/tex]
[tex]t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084[/tex]
Rounding off we get,
t = 2.08 , t = 0.42
Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds
