Differential equation for velocity

i was working on a differential equation to find the initial velocity of a rocket to get to height h.

so i have, which i got and its correct cause its the solution in the book, v^2 = 2g(R/(R+h))-2gR+V(0)^2

v = velocity
g = acceleration of gravity at earths surface
R= radius of earth
h = height
and i assume v(0) is the velocity v at height=0 so v(0)=0?

Problem is when i keep getting a negative number. so i assume I'm making an amateur mistake.

What am i doing wrong? ive tried converting etc. im using km

The initial velocity cannot be 0. Unless how are you gonna send the rocket up? It is velocity at the maximum height that should be zero.

so should i be trying to solve for v(0) instead of v? i guess you're saying the velocity goes to 0 when we reach the height we want to achieve right?

That's right. In your question it self has asked to find the initial velocity hasn't it?

Yes, to get to a certain height. i will work on it, then post what i have.

v^2 = 2g((R^{^2})/(R+h))-2gR+V(0)^2

v₀^2 = 2gR - 2g((R^{^2})/(R+h)) + v^2

so v^2 is 0 cause at the velocity of h the rocket stops. and now solve for V₀

v₀ = √(2gR) - √(2g((R^{^2})/(R+h)))

R= 6378.1 km
h= 450 km
g= 9.80665 m/s² = 127094.184 km/hr²

v₀ = √((2)(127094.184)(6378.1)) - √(2(127094.184)((6378.1^{^2})/(6378.1+450)))

v₀ = 1328.1287 km/hr = 368.9246 m/s


Did i solve this correctly?

when i try to do this in metres instead of kiometers using the variables
R= 6378100 m
h= 450000 m
g= 9.80665 m/s²

i get v₀ = 374.7107 m/s

is this more or less accurate?

Now your main equation is,

right...

with v=0, we have,



not

OMG you are right! i feel a fool. im getting matching answers now. of 2.8713 km/s and 2871.2917 m/s