Question

I need to find out how long it takes him to hit the water

Answer 1

Answer:

1.26

Step-by-step explanation:

When he hits the water, the height will be 0. Substitute h=0.

$0 = -16t^2+2t+23$

Solve using the quadratic equation.

$x = \frac{-b+/-\sqrt{b^2-4ac} }{2a}$

Substitute a = -16, b=2, and c=23.

$x = \frac{-b+/-\sqrt{b^2-4ac} }{2a}\\\\x = \frac{-2+/-\sqrt{(-2)^2-4(-16)(23)} }{2(-16)}\\\\x = \frac{-2+/-\sqrt{4+1472} }{-32}\\\\x = \frac{-2+/-\sqrt{1476} }{-32}\\x = \frac{-2+/-38.4}{-32}\\x = \frac{-2+38.4}{-32}=\frac{36.6}{-32}=-1.14\\or\\x = \frac{-2-38.4}{-32}=\frac{-40.4}{-32}=1.26$

Since time in seconds must be positive then 1.26 is the solution.