Thread: 3 second burn rocket problem

Pretty sure I have every part of this problem solved but my final answer concerns me as it seems a little high.

for 0<=t<=3
and for t>3

I should find the maximum altitude and time in which it gets there. My answer is 56s and 18565.18meters.

**additional info**
initial velocity ( is 366.4m/s)
initial position ( ) is 0

after taking integral of v(t)
**
= position fxn until 3 seconds.
after 3 seconds of burn

=position fxn after 3s initial burn.

at t=56.2145s

where t=56.2145s


17087.98+1477.2=18565.18meters

Is it me or is that pretty dang high?

I guess I don't need to add the two x positions since the second one already has it in there. So 17087.98meters.

for 0<=t<=3
and for t>3
So,
initial velocity ( is 366.4m/s)
initial position ( ) is 0

after taking integral of v(t)

Then, at 3 seconds:

and


so after 3 seconds






Finally:
at t=53.23seconds
then


So, it reaches a height of 15479.02m after flying for 53.23s. Still seems high to me...

For some reason I switched from 9.8 to 9.88 as grav. constant. I adjusted them on my paper but don't have the time to do it again here. The values are very nearly the same anyway.