Question

A projectile is fired vertically upwards and can be modeled by the function h(t)= -16t to the second power+600t +225 during what time interval will the project I’ll be more than 4000 feet above the ground round your answer to the nearest hundredth

Answer 1

Given:

`h(t)=-16t^2+600t+225`

To find the time interval when the height is about more than 4000 feet:

Let us substitute,

`\begin{gathered} h(t)\ge4000 \\ -16t^2+600t+225\ge4000 \\ -16t^2+600t+225-4000\ge0 \\ -16t^2+600t-3775\ge0 \end{gathered}`

Using the quadratic formula,

Here, a= -16, b=600, and c= -3775

`\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-600\pm\sqrt[]{600^2-4(-16)(-3775)}}{2(-16)} \\ =\frac{-600\pm\sqrt[]{360000^{}-241600}}{-32} \\ =\frac{-600\pm\sqrt[]{118400}}{-32} \\ =\frac{-600\pm40\sqrt[]{74}}{-32} \\ =\frac{-75\pm5\sqrt[]{74}}{-4} \\ t=\frac{-75+5\sqrt[]{74}}{-4},x=\frac{-75-5\sqrt[]{74}}{-4} \\ t=7.99709,t=29.5029 \end{gathered}`

So, the interval is,

`8.00\le\: t\le\: 29.50`