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PilotLPTM [1.2K]
3 years ago
13

The height reached by the rocket depends on the amount of fuel

Mathematics
2 answers:
emmasim [6.3K]3 years ago
4 0
Initial displacement = dboost = v0t + 1/2at^2 
<span>hboost = dboost sin theta </span>
<span>xboost = dboost cos theta </span>
<span>vboost = v0 + at </span>
<span>vxboost = vboost cos theta </span>
<span>vyboost = vboost sin theta </span>

<span>To get the maximum altitude, use conservation of energy </span>
<span>initial PE + vertical KE </span>
<span>= mg (hboost) + 1/2 m vyboost^2 </span>
<span>= final PE = mg (hfinal) </span>

<span>hfinal = hboost + vyboost^2 / 2g </span>
<span>= hboost + ((v0 + at) * sin (theta))^2 / 2g </span>

<span>b) flight time = </span>
<span>boost time + time to rise ballistically from hboost to hfinal </span>
<span>+ time to fall from hfinal </span>

<span>The time to fall a distance h can be gotten by: </span>
<span>h = 1/2 gt^2, so t = sqrt (h/2g) </span>

<span>So flight time = t + sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g) </span>

<span>Horizontal range = xboost + vxboost * time of ballistic flight </span>

<span>= (v0t + 1/2at^2) cos theta </span>
<span>+ (v0 + at) cos theta * (sqrt ((hfinal - hboost)/2g)+ sqrt (hfinal/2g)) </span>

<span>Lots of plugging and chugging to do now. </span>

<span>EDIT--the problem said acceleration was constant, so we don't have to worry about loss of mass during the boost. Realistic? No. But that's what the problem says. Once the thing is ballistic, it's a moot point anyway

</span>
gavmur [86]3 years ago
3 0
True as u cant have a rocket flying at a high height if its low on fuel as if the fuel runs out mid flight they will just crash

brainliest:))))))))


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Step-by-step explanation:

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