Hello all, I was wondering if someone could please help me with a problem. It is regarding the law of universal gravitation, this time using conservation of mechanical energy. We know that F = GMm/R^2 **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 27% of the escape speed.** I approached this problem two ways and have stopped in the middle, because I don't know how to continue. I was wondering if someone can help me find h or the height in meters. First thing I tried: .5mv^2 - GmM/Radius of earth = .5mv^2 - GmM/h 0 = GmM/Radius of earth - GmM/h goes to h = Radius of earth I stopped here Second thing I tried: .5mv^2- GmM/Radius of earth = .5m(.27v)^2- GmM/Radius of earth + h I do not know how to manipulate this problem to get radius of earth + h Thank you for your assistance!