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**A particle is thrown away with degree elevation of α above the earth surface. Whenever the particle will undergo motion which is going farther from the initial point if and only if the vector components of its rate of motion are parallel to the vector components of its vector of position. Let g be the earth’s gravitational acceleration. What will be the value of α in order for the particle to always undergo motion which is always going farther from the initial point?**

My friend and I have 2 different answers. I answer

**tan α = (y+1/2gt**based on parabollic motion. I assume parabollic motion as an approach because every time, the particle will always go farther from the initial point with degree elevation α. It won't be vertical motion because at the minimum point, the particle will go down again and go back to its initial point. Thus, I use parabollic motion formulas to find α with y, g, and x variable. On the other hand, my friend answers

^{2}/ x)**sin α > sqrt(8/9)**based on his vector analysis which I don't understand *haha lol*.

**What's your opinion about the answers? Or do you have any different solution to this problem? Help me please :)**