Find the maximum angle above the horizontal with which an object can be projecte

Question

Find the maximum angle above the horizontal with which an object can be projected so that its distance from the launch point always increases, that is, as t increases, the distance from the launch point also increases. For example, at small angles like 0 degree or 1 degree it is obvious that the projectile always moves away from the launch point (at least until it bits the ground) but at large angles like 90 degree or 89 degree although the projectile initially moves further away, eventually it turns around and hits or nearly hits the launch point, so clearly these angles are too big. You may leave your answer in terms of inverse trig functions (e.g., arctan( 1/3)), but since you have a calculator handy, an answer in degrees would also be useful. As always, you will be graded on your solution, not just the answer. Show your work. You will save yourself some effort if, instead of using the "distance from the launch point," you use that distance squared. This saves you from differentiating a root and works because a nonnegative function and its square are monotonic in each other.

Explanation / Answer

The distance traveled horizontally by a projectile launched at a certain angle (call it alpha) from the horizontal has the formula:

x=v0tcos(alpha) (1)

Suppose v0=constant.

The question is strange. x always increases with t, as eq(1) shows, for any angle between 0 and 180 deg (for alpha between 90 and 180 deg, the projectile moves on the negative part of Ox axis, but the formalism is the same, and perfectly possible). Of course the projectile will come back to the earth, sooner or later, if its velocity is not too high.

So, I don't understand the question.

OBS: If you're talking about the condition, for a projectile, to become satellite, this is not about the angle, but about its velocity.

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