Projectile Motion: Escape Velocity and Altitude at 55%

[SoccaCrazy24]

Question 1: If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is 96% of the escape speed?
On this question i understand i am not finding how high it would go if it were launched at 96% of the escape velocity, but rather that when it gets to 96% of the escape speed. I am unsure of how I am to find this... and I would like to atleast get a concept for this one... which will help me with question 2 as well. I believe it will deal with energy conservation, but i cannot think of a way that i can involve velocity with this. Can I get atleast an intuitive way to help lead me to get somewhere with this? If I can get this question I can surely get question 2.
Question 2: A projectile is launched vertically from the surface of the Moon with an initial speed of 2300 m/s. At what altitude is the projectile's speed 55% of its initial value?

 

[mathphys]

Gravity will act on the projectile. So what is the formula for its final velocity in terms of t, its initial velocity and the acceleration?

 

[skeeter]

if a projectile is launched from Earth at escape speed, then its total energy is equal to 0 ...

K + U = 0

from this relationship, escape speed is calculated as v = sqrt(2GM/R)
where G = grav. constant, M = Earth mass, and R = Earth radius.

I'm assuming that you already know all the above, but the point is this ...
during its entire trip, total energy remains a constant equal to zero.

let v = escape velocity, k = .96, and r = variable distance from Earth's center

(1/2)m(kv)^2 - GMm/r = 0

since escape velocity v = sqrt(2GM/R)

(1/2)m*k^2*2GM/R - GMm/r = 0

solving the above equation for r yields ...

r = R/k^2

where r = distance from the Earth's center when it is at 96% of escape speed