I finding the distance and the time

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SUMMARY

The discussion focuses on calculating the height and total time of a bottle rocket's flight, which accelerates upward at 30 m/s² for 0.5 seconds. The maximum height achieved is 15.23 meters, derived from the initial upward acceleration and subsequent calculations. The velocity at the end of the acceleration phase is determined to be 15 m/s. The participants clarify that the acceleration during descent differs from the ascent, and the total time from launch to landing remains to be calculated.

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Homework Statement



1. A bottle rocket has an engine that accelerates it upward. Then it coasts to its highest point before returning to the ground. Assuming that the engine creates a net acceleration of 30 m/s2 for 0.5 s, how high does it go, and how long does it take to reach the ground again?

acceleration= 30m/s^2
time=0.5s

Homework Equations


v^2-u^2=2ax

The Attempt at a Solution


0^2-(15m/s)^2=2(-9.8m/s^2)x
-225=-19.6x
-255/-19.6
x=11.48m
to get total height i did 11.48m+3.75m=15.23m

now how do i get velocity for the other question
v=at
v=30m/s^2*0.5s
v=15m/s
does this work?
 
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The acceleration isn't 30m/S^2 on the way down. Nor is the time 0.5 seconds.
 
Mary0524 said:
...
15.23m
That's what I come up with.
now how do i get velocity for the other question
v=at
v=30m/s^2*0.5s
v=15m/s
does this work?
I thought the second part of the question was "find: Total time, from launch to landing"
?
 

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