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I have been struggling with this problem for a whole day, and finally admitting that i can't solve it, so i ask for help

I am familiar with the escape velocity , and the speed needed to brake into interplanetary space - the newton cannonball thought experiment. But i was wonder about launching a cannonball perpendicular to the gravitational field - vertical lunch. Ignoring air resistance, and other celestial bodies , what is the initial speed the cannonball must have in order to break trough gravitational influence of the earth and head to infinity?

I tried , but i got just mathematical understanding of calculus, i never learned it's applications in physics.

I know only that this values change over time like this:

v(t) = Vo-g*t g= M*G/r^2 , and R=ro+V(t) H= Vo*t-(g*t^2)/2 or after brief calculation, H=(Vo^2)/2*g

where M is the mass of earth

r the radius

t-time

ro initial semi major axis of earth

vo- initial velocity

H-max height in homogeneous g-field

because speed tends to reduce for g each second , and g tends to reduce by inverse square of r

there should , in my opinion be a sufficiently large initial velocity with which a verticaly launched projectile would leave the earth to infinity, as R-> infinity g ->0 , LimV= Vmin and LimH=infinity

How to mathematically express that kind of solution, if there is one?

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# Vertical projectile mathematics

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