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Question Number 182874 by peter frank last updated on 15/Dec/22

	Answered by TheSupreme last updated on 16/Dec/22

	$V = v c o s (θ) \vec{x} + (v s i n (θ) - g t) \vec{y}$ $S = v c o s (θ) t \vec{x} + (v s i n (θ) t - \frac{1}{2} {g t}^{2})$ $e q o f p l a n e$ $y = t a n (α) x$ $i f p a r t i c l e h i t s h o r i z z o n t a l y$ $v s i n (θ) - g t = 0$ $t = \frac{v s i n (θ)}{g}$ $S^{*} = \frac{v^{2}}{g} s i n (θ) c o s (θ) \vec{x} + \frac{1}{2} \frac{v^{2} {s i n}^{2} (θ)}{g} \vec{y}$ $y = t a n (α) x$ $\frac{1}{2} \frac{v^{2} {s i n}^{2} (θ)}{g} = \frac{v^{2}}{g} s i n (θ) c o s (θ) t a n (α)$ $\frac{1}{2} t a n (θ) = t a n (α)$ $t a n (θ) = 2 t a n (α)$ $d = \sqrt{1 + {t a n}^{2} (α)} x = \frac{x}{c o s (α)} = \frac{v^{2} s i n (θ) c o s (ϑ)}{g c o s (α)}$ $. . .$ ${t a n}^{2} (θ) = 4 {t a n}^{2} (α) \to \frac{1}{{c o s}^{2} (θ)} = 1 + 4 {t a n}^{2} (α)$ ${c o s}^{2} (θ) = \frac{1}{1 + 4 {t a n}^{2} (α)}$ ${s i n}^{2} (θ) = \frac{4 {t a n}^{2} (α)}{1 + 4 {t a n}^{2} (α)}$ $. . .$ $d = \frac{v^{2}}{g} \frac{2 t a n (α)}{1 + 4 {t a n}^{2} (α)} \frac{1}{c o s (α)}$
	Commented by peter frank last updated on 17/Dec/22

	$thank you$
	Commented by peter frank last updated on 31/Dec/22

	$diagram please ?$ $where this came from$ $↓↓↓↓$ $d = \sqrt{1 + {t a n}^{2} (α)} x = \frac{x}{c o s (α)} = \frac{v^{2} s i n (θ) c o s (ϑ)}{g c o s (α)}$
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