Question

a model rocket is launched from ground level into the air. its height is y, in feet, after x seconds can be represented by the equation y=160x-5x sq. what is the elapsed time in seconds, from the time the rocket is launched until it reaches ground level again?

Answer 1

Ground level is 0 so Y = 0. Now what? :/

Answer 2

Ok, first of all, make it clear to me:\[y=160x-5x ^{2}\]Is this the equation they gave you?

Answer 3

Yes

Answer 4

I need help to answer the question and I also would like to know what -5X^2 means. Is it when the rocket starts to descend?

Answer 5

Then you must derivate that equation two time in order to find acceleration.\[\frac{ d }{ dx }y=160-10x=v \rightarrow \frac{ d }{ dx }\frac{ d }{ dx }y=\frac{ d }{ dx }v=-10=a\]With v calculate the velocity when x is zero.\[160-10(0)=160=v _{o}\]Now use that acceleration and the initial velocity to find the time. \[x _{f}=x _{o}+v _{o}t+\frac{ 1 }{ 2 }at ^{2} \rightarrow x _{f}=0 \rightarrow t=\frac{ -v _{o} \pm \sqrt{v ^{2}_{o}-4(\frac{ 1 }{ 2 }a})(x _{o}) }{ 2(\frac{ 1 }{ 2 }a) } \rightarrow x _{o}=0 \rightarrow \]\[t=\frac{ -160\frac{ m }{ s } \sqrt{(160\frac{ m }{ s })^{2}} }{ -10\frac{ m }{ s ^{2} } }=2560s\]

Answer 6

I don't know how to answer it in another way.

Answer 7

I do not understand that way in solving it. I am only in algebra 1. I also do not know physics.

Answer 8

@satellite73

Answer 9

Oh no I got it! Wait

Answer 10

You just have to solve for when y=0\[y=160x-5x ^{2} \rightarrow y=0 \rightarrow 0=160x-5x ^{2}\]\[x=\frac{ -160\pm \sqrt{160^{2}-4(-5)(0)} }{ 2(-5) })=\frac{ -160\pm 160 }{ -10 }=0o32\] The answer is 32

Answer 11

Do you get what happened between the first and second formula?