Solve Rocket Problem with Step-by-Step Guide | Imaginary Number Solution

[takelight2]

Homework Statement

A toy rocket moving vertically upward passes by a 2.1 m -high window whose sill is 8.0 m above the ground. The rocket takes 0.13 s to travel the 2.1 m height of the window.


What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.

Relevant Equations

s = ut + 1/2at^2

This is what I tried and it makes perfect sense to me. When i plug in the numbers to what I ended up with, I get an imaginary number for U initial...

Work is attached below.

 

[vela]

One of the equations you wrote down, $v_0 = u^2+2gh$, is incorrect. You can see that just from comparing the units on the lefthand side with the units of the righthand side.

 

[takelight2]

One of the equations you wrote down, $v_0 = u^2+2gh$, is incorrect. You can see that just from comparing the units on the lefthand side with the units of the righthand side.

Oh yeah that makes sense. So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?

 

[TSny]

So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?

When you wrote the equation $v_0 = u^2 + 2gh$, not only is there the mistake of not squaring $v_0$, but there is also a sign error in the equation. Can you spot it?

 

[Lnewqban]

Oh yeah that makes sense. So that would change my final equation to this:

u = sqrt[([(s/t)+(1/2)(g)(t)]^2) -2gh]

But it still is giving me the wrong value D: Any other ideas?

It seems to me that, if you are assuming u to be the launching velocity and Vo the velocity of the rocket when it was by the top of the window, either both signs in that equation are incorrect, or the input value of g should be negative.
Since the rocket is decelerating while flying upwards, Vo < u.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot4