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velouria131

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## Homework Statement

A projectile is launched with a speed v at an angle theta above the horizontal. Ignore air resistance. Derive an expression for the angle theta in terms of the parameters of the problem such that the horizontal distance from the launch of the object is N times greater than the maximum height achieved during its flight. Assume that the object lands at the same vertical position from which it was launched and that the acceleration due to gravity is taken as g

## Homework Equations

v = vo + at

vav = ½(vo + v)

x = xo + ½(vo + v)t

x = xo + vot + ½at²

v² = vo² + 2a(x - xo)

x = xo + vt - ½at²

## The Attempt at a Solution

First of all, thanks for any help and hello. I have sat here for quite awhile deliberating, and I seem to just convolute any potential solutions. The first thing I did was draw vector R at angle theta from the horizontal. I then made two tables; table X and table Y. I defined the initial velocity in the X direction as vcos? and the initial velocity in the Y direction as vsin?. I defined the acc. in the Y direction as -9.8 m/s^2. I assumed from this point that I would have to construct an equation - inverse tangent vsin? / vcos?, in terms of vsin? and vcos? that would end of being numerical. I cannot for the life of me figure this part out. The key seems to be that the horizontal distance traveled is N times the maximum vertical distance, which would be N times the the distance traveled by the object whose finally velocity is zero and initial velocity vsin?. Where do I go from here? Or am I even on the correct track?

Thanks again.

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