We are given an equation that computes the height of a projectille as a function of time t:
h = -16t² + 160t + 29
In order to calculate the time at which the projectile will be 173 feet above the ground, first we replace 173 for h:
173 = -16t² + 160t + 29
Now, subtract 173 on both sides, so we get 0 on the left side of the equation:
0 = -16t² + 160t + 29 - 173
0 = -16t² + 160t - 144
In order to solve for t from the above expression, we can use the quadratic formula:
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a, b, c are the coefficients of the second-order polynomial of the form
0 = at² + bt + c
In this case, a equals -16, b equals 160 and c equals -144.
By replacing these values into the above formula, we get:
[tex]\begin{gathered} t=\frac{-160\pm\sqrt[]{160^2-4\times(-16)\times(-144)}}{2\times(-16)} \\ t=\frac{-160\pm128}{-32} \\ t=\frac{32}{32}=1 \\ t=\frac{288}{32}=9 \end{gathered}[/tex]The quadratic formula gives us two possibles solutions, t = 1 or t = 9, then the projectile will be 173 feet above the ground after 1 second.
