Question 970516
Assumptions: 

1) The rocket is starting on the ground (and not on a building or hill)

2) The rocket comes back to the same ground level (same height) as the starting point.

3) Dealing with wind/air resistance makes this problem a lot more complicated, so we will ignore those factors.

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{{{h = -(g/2)t^2 + v[0]*t + h[0]}}} is the height formula


h = height at time t (where both h and t are nonnegative)

g = acceleration of gravity at sea level (roughly 32 ft/sec^2)

{{{v[0]}}} is the initial velocity

{{{h[0]}}} is the initial height



{{{h = -(g/2)t^2 + v[0]*t + h[0]}}}


{{{h = -(32/2)t^2 + 160*t + 0}}}


{{{h = -16t^2 + 160t}}}


The rocket's max height is when {{{t = -b/(2a) = -160/(2*(-16)) = 5}}} seconds


Question: How long will it take the rocket to reach maximum height? 
Answer: 5 seconds


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The rocket hits the ground at height h = 0.


{{{h = -16t^2 + 160t}}}


{{{-16t^2 + 160t = h}}}


{{{-16t^2 + 160t = 0}}} Plug in h = 0


{{{-16t(t - 10) = 0}}}


{{{-16t = 0}}} or {{{t - 10 = 0}}}


{{{t = 0}}} or {{{t = 10}}}


The rocket is on the ground at {{{t = 0}}} or {{{t = 10}}} seconds. The first time value is the starting point whereas the second time value is the ending point.


So the rocket starts on the ground at t = 0 seconds, flies up, then comes back down and hits the ground at t = 10 seconds. The total flight time is 10 seconds.



Question: How long would it take the rocket to return to earth from the time it was launched?
Answer: 10 seconds