Question-182874

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)}

\dots

{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)}

\dots

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)}