Question

14.) Suppose h = -16t2 + 160t + 29 computes the height h, abovethe ground in feet, of a projectile fired upward in t seconds. SOLVEthe equation above for the exact time t when the projectile will be 173feet above the ground?(12 pts)

Answer 1

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:

`t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

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:

`\begin{gathered} t=\frac{-160\pm\sqrt[]{160^2-4*(-16)*(-144)}}{2*(-16)} \\ t=(-160\pm128)/(-32) \\ t=(32)/(32)=1 \\ t=(288)/(32)=9 \end{gathered}`

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.