(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Suppose we attempt to account for air resistance in our projectile motion in the following (incorrect) way: we alter g so that the acceleration in the y direction is -8 m/s^{2}, and introduce a horizontal acceleration of -3 m/s^{2}. With these changes, find the landing point of a projectile fired with initial speed 32 m/s at an angle of 25°.

2. Relevant equations

x - x_{0}= (v_{0}cos theta_{0})t + 1/2 at^{2}(note this is changed from the textbook definition- by the addition of the term (+ 1/2 at^{2})- bc of the introduction of horizontal acceleration)

y - y_{0}= (v_{0}sin theta_{0})t + 1/2 at^{2}

v_{y}= v_{0}sin theta_{0}+ at

v_{y}^{2}= (v_{0}sin theta_{0})^{2}+ 2a(y - y_{0})

and other constant acceleration equations altered for the purposes of projectile motion

for constant acceleration:

R = (2v_{0}^{2}/g) (sin 2theta_{0})

where R = the horizontal range of the projectile

3. The attempt at a solution

I was wondering if there was a way to solve this without assuming the landing point is the same as the launch point (constant elevation.) I solved it assuming constant elevation, but as it doesn't say that, I don't know if that was perhaps a bad assumption.

With my assumption, I found that the projectile's landing point was 80.9 m from its starting point.

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# Projectile Motion

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