Kinematics of a falling rocket

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


A physics student launches a model rocket from the ground at t=0. The rocket ascends with a constant vertical acceleration of a1= 2.5m/s^2 until T1= 5.0s, when the acceleration increases to a2= 5.0m/s^2 until T2= 10.0s. When t>T2, the engine shuts down entirely.
(b) What is the velocity of the rocket when it hits the ground?

Homework Equations


kinematic equations


The Attempt at a Solution



v(T1)= 0+(2.5)(5)=12.5m/s

v(T2)=12.5+(5)(5)=37.5m/s

xT1=0+0+1/2at^2=1/2(2.5)(5)^2=31.25m

xT2=31.25+12.5(5.0)+(1/2*5*5^2)=156.25m

V^2=37.5^2-2(9.8)(156.25)=-40.5m/s

but the answer is -66.8 m/s

what am i doing wrong?

Thanks for your help!
 

Answers and Replies

  • #2
38
0


everything is right until the last line where you use v^2 = u^2 + 2as. remember to define a positive direction!!
 
  • #3
16
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so v^2=37.5^2+2(-9.8)(-156.25m)
v=40.7m/s

..:( still didn't give me -66.8 m/s

positive direction is up
 
  • #4
38
0


haha
either you forgot to square root
or u need to get a new calculator!! =P
 
  • #5
38
0


nvm i just realised u did take the square root
maybe u need a new calculator =DD
but most likely you made a typing error
try type it in again? it comes out on my calculator
 
  • #6
16
0


ohhhhh gosh finally got it on my calculator >.<


Thanks for helping! :)
 
  • #7
38
0


LOL no worries xDD i know how frustrating those damn calculators can be too >=[
 

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