What-is-the-maximum-angle-to-the-horizontal-at-which-a-stone-can-be-thrown-and-always-be-moving-away-from-the-thrower-

Question Number 18261 by Tinkutara last updated on 17/Jul/17

What is the maximum angle to the

horizontal at which a stone can be

thrown and always be moving away

from the thrower ?

Answered by ajfour last updated on 26/Jul/17

r = \sqrt{x^{2} + y^{2}}, x = u_{x} t, y = u_{y} t - \frac{{gt}^{2}}{2}

\frac{dr}{dt} = \frac{(\frac{2 xdx}{dt} + \frac{2 ydy}{dt})}{2 \sqrt{x^{2} + y^{2}}}

= \frac{u_{x}^{2} t + (u_{y} t - \frac{{gt}^{2}}{2}) (u_{y} - gt)}{1 \sqrt{x^{2} + y^{2}}}

\tan θ = \frac{u_{y}}{u_{x}}

\frac{dr}{dt} > 0 \Rightarrow t [u_{x}^{2} + (u_{y} - \frac{gt}{2}) (u_{y} - gt)] > 0

or \frac{g^{2} t^{2}}{2} - \frac{{3 u}_{y} gt}{2} + u_{x}^{2} + u_{y}^{2} > 0

\Rightarrow D = \frac{{9 u}_{y}^{2} g^{2}}{4} - {2 g}^{2} (u_{x}^{2} + u_{y}^{2}) < 0

\Rightarrow \frac{u_{y}}{\sqrt{u_{x}^{2} + u_{y}^{2}}} < \frac{2 \sqrt{2}}{3}

or \frac{u_{y}}{u} < \frac{2 \sqrt{2}}{3} \Rightarrow \sin θ < \frac{2 \sqrt{2}}{3}

\tan θ < 2 \sqrt{2} \cos θ > \frac{1}{3}

θ < \cos^{- 1} (1 / 3) which is same as

θ < \tan^{- 1} (2 \sqrt{2}) .

Commented by Tinkutara last updated on 26/Jul/17

Thanks Sir!