feet as a function of time

[RPMACS]

A missile is fired vertically into the air. The distance s (in feet) above the ground as a function of time t (in seconds) is given by the equation below.
(a) When will the missile hit the ground?
(b) When will the missile be 1000 feet above the ground?

Help!

 

[galactus]

The equation below?.

Use PREVIEW to make sure you are posting what you are meaning to post.

Whatever the equation is, the missile will hit the ground when its height is 0. Set your quadratic equal to 0 and solve for t. There will be two solutions. One will more than likely be extraneous. To find how long until its 1000 feet in thre air, just set the equation equal to 1000 and solve for t.

 

[RPMACS]

Sorry, I forgot the equation...

h= 300+500t-16^2

+/- (in square) -h+1275 +35/4

is this close?

Thanks

 

[soroban]

Hello, RPMACS!

This requires only basic Algebra.
Exactly where is your difficulty?

A missile is fired vertically into the air.

The distance \(\displaystyle h\) (in feet) above the ground as a function of time \(\displaystyle t\) (in seconds) is given by the equation:

. . \(\displaystyle h \:=\:300 + 500t - 16t^2\)

(a) When will the missile hit the ground?

"Hit the ground" means \(\displaystyle h \,=\,0.\)

\(\displaystyle \text{We have: }\:300 + 500t - 16t^2 \:=\:0 \quad\Rightarrow\quad 4t^2 - 125t - 75 \:=\:0\)

\(\displaystyle \text{Quadratic Formula: }\;t \;=\;\frac{\text{-}(\text{-}125) \pm\sqrt{(\text{-}25)^2 - 4(4)(\text{-}75)}}{2(4)} \;=\;\frac{125 \pm\sqrt{23,\!125}}{8} \;=\;\frac{125\pm25\sqrt{37}}{8}\)

\(\displaystyle \text{Therefore: }\;t \;=\;34.63363291 \;\approx\;34.6\text{ seconds.}\)

(b) When will the missile be 1000 feet above the ground?

\(\displaystyle \text{We are told that }h \,=\,1000\)

\(\displaystyle \text{We have: }\;300 + 500t - 16t^2 \:=\:1000 \quad\Rightarrow\quad 4t^2 - 125t + 175 \:=\:0\)

\(\displaystyle \text{Quadratic Formula: }\;t \;=\;\frac{\text{-}(\text{-}125) \pm\sqrt{(\text{-}125)^2 - 4(4)(175)}}{2(4)} \;=\;\frac{125 \pm\sqrt{12,\!825}}{8} \;=\;\frac{125 \pm15\sqrt{57}}{6}\)

\(\displaystyle \text{Therefore: }\;t \;=\;\begin{Bmatrix}\frac{125 - 15\sqrt{57}}{8} & \approx & 1.47 \\ \\[-3mm] \frac{125 + 15\sqrt{57}}{8} & \approx & 29.78 \end{Bmatrix}\,\text{ seconds}\)

 

[RPMACS]

I have problems reducing from the quadratic formula. Your explanation/break down really helps.

Thanks for your help!