Maximum range of projectile launched from elevation

classical-mechanicsintuitionkinematicsphysicsprojectile motion

I am trying to find the maximum range of projectile from an elevation. I found the answer in this question, but I have two questions:

  1. Why does $y$ need to be $0?$
  2. Why do they differentiate with respect to $\theta,$ and what is the meaning of this?

Best Answer

  1. Why does $y$ need to be $0?$

Because the projectile's range $R$ is its horizontal distance travelled, which is precisely its $x$-coordinate at its landing point, that is, when there is no elevation, that is, when $y=0.$

  1. Why do they differentiate with respect to $\theta,$ and what is the meaning of this?

The projectile's range $R$ varies according to its launch angle $\theta,$ so is maximised when $\frac{\mathrm dR}{\mathrm d\theta}$ equals zero. (Consider the graph of $R$ against $\theta;$ $R$ attains its maximum value when its gradient $\frac{\mathrm dR}{\mathrm d\theta}$ equals zero.)

P.S. I am the author of the cited Question; even though I accepted Patrick's Answer, please treat my self-Answer there as the natural continuation of the work that I'd started in the Question.

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