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Rocket Velocity

  • Thread starter Jimbo57
  • Start date
  • #1
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Homework Statement


During the first 50 s of a rocket's flight, the rocket is propelled straight up, so that in t seconds its height s is s(t) = t^3/sqrt 10 ft.

a) How high does the rocket travel in 50 s?
b) What is the average velocity of the rocket during the first 50 s?
c)What is the average velocity of the rocket during the first 125 ft?

Homework Equations


s(t) = t^3/sqrt 10 ft
v(t) = 3t^2/sqrt 10

The Attempt at a Solution



a) Pretty straightforward: s(50)= 50^3/sqrt 10
= 125000/sqrt10
≈39,528.47 ft
b) average velocity = distance traveled/time elapsed
= 39528.47/50
≈790.57ft/s (I have a feeling this may be wrong)
c) 125ft=t^3/sqrt10
t=cbrt125*sqrt10
= 7.34s

Average = 125ft/7.34s
= 17.03ft/s (I also think this may be wrong)

If I was an error, where would I be? I feel like a rocket traveling at 17ft/s after 7 seconds may be a tad on the low side.
 

Answers and Replies

  • #2
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Bump! Any takers?
 
  • #3
Filip Larsen
Gold Member
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The calculations look good to me.

Since this rocket accelerates straight up with an acceleration of a = 6t/sqrt(10) the thrust of the rocket must be a + 1g, than is, a thrust of 1 g at t = 0 which slowly builds up to 2 g after 17 seconds. You are correct that this is "a tad on the low side" and that a real rocket would be designed to have an initial acceleration a good amount larger than 1 g.
 
  • #4
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Looking at this in terms of acceleration put things into perspective to me, and made my calculations look more reasonable. Thanks Filip!
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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Bump! Any takers?
Your work looks good.

You should be aware that the rules of this Forum require you to wait 24 hours before bumping. I see you did wait about half that. ... not too bad, but ...
 
  • #6
96
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My apologies Sammy, I have to admit that I haven't read any of the rules :( Won't happen again!
 

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