Question

Hello, stuck on this practice problem. I also need to show all work.

Answer 1

Given the model for the height of a ball:

`h(t)=-16t^2+5t+15`

To hit the ground, the height must be h(t) = 0. Then:

`\begin{gathered} -16t^2+5t+15=0 \\ \Rightarrow16t^2-5t-15=0 \end{gathered}`

We use the general solution for a quadratic equation:

`ax^2+bx+c=0\Rightarrow x=(-b\pm√(b^2-4ac))/(2a)`

From the problem, we identify:

`\begin{gathered} a=16 \\ b=-5 \\ c=-15 \end{gathered}`

Using the formula:

`\begin{gathered} t=(5\pm√(25+4\cdot15\cdot16))/(2\cdot16)=(5\pm√(985))/(32) \\ \\ \Rightarrow t_1=1.137 \\ \Rightarrow t_2=-0.825 \end{gathered}`

We only take the positive value (because the time is always positive!). Then, the answer is:

`1.137\text{ seconds}`