Question

prove that the path of the projectile is parabolic. also find 1. maximum height 2. time of flight
3.horizontal range..​

Answer 1

Explanation:

let the speed of the projected object = to (a)

vertical component =

$a \sin( \alpha )$

horizontal component

$a \cos( \alpha )$

horizontal acceleration = 0

vertical acceleration = -g

$use \: this \: eqution \: to \: get \: 2 \: equtions \: \\ vertically \: and \: horiozontally$

$s = ut + \frac{1}{2} a{t}^{2}$

therefore we get for horizontal

$x = a \cos( \alpha ) t+ \frac{1}{2}a {t}^{2} \\ x = a \cos( \alpha ) t \\ then \: get \: an \: equatin \: for \: t \\ \frac{x}{ \cos( \alpha ) } = t$

for vertical

$y = u \sin( \alpha ) t + \frac{1}{2} a {t}^{2} \\ then \: subsitude \: eqution \: we \: get \: for \: \\ (t) \\y = \\ a \sin( \alpha ) \times \frac{x}{a \cos( \alpha ) } - \frac{1}{2} \times \frac{ {x}^{2} }{ {a}^{2} {cos}^{2} \alpha }$

$y = \tan( \alpha ) x - \frac{1}{2}g \frac{ {x}^{2} }{ {a}^{2} {cos}^{2} \alpha }$

this equatio is equal to the eqaution of a parabola

$y = a {x}^{2} + bx + c \\ where \: a \: b \: c \: are \: constants$

therefore we proove that the projectile is parabolic