Answer:
1.26
Step-by-step explanation:
When he hits the water, the height will be 0. Substitute h=0.
[tex]0 = -16t^2+2t+23[/tex]
Solve using the quadratic equation.
[tex]x = \frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
Substitute a = -16, b=2, and c=23.
[tex]x = \frac{-b+/-\sqrt{b^2-4ac} }{2a}\\\\x = \frac{-2+/-\sqrt{(-2)^2-4(-16)(23)} }{2(-16)}\\\\x = \frac{-2+/-\sqrt{4+1472} }{-32}\\\\x = \frac{-2+/-\sqrt{1476} }{-32}\\x = \frac{-2+/-38.4}{-32}\\x = \frac{-2+38.4}{-32}=\frac{36.6}{-32}=-1.14\\or\\x = \frac{-2-38.4}{-32}=\frac{-40.4}{-32}=1.26[/tex]
Since time in seconds must be positive then 1.26 is the solution.
