Question

A)Show that the path of projectile is parabola. (2 ) B) derive the formula of (1)time of flight. ( 1) (2) Maximum height (1) (3) horizontal range. (1)


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Answer 1

Answer:

(A) Path of projectile is parabola

$x = u \cos( \alpha ) t \: \: \: \: \\ t = \frac{x}{u \cos( \alpha ) } \: \: \: \: \: \: eq \: 1\\ y = u \sin( \alpha ) t - \frac{1}{2}g {t}^{2} \\ from \: eq \: 1 \\ y = u \sin( \alpha ) (\frac{x}{u \cos( \alpha ) }) + \frac{1}{2}g ({ \frac{x}{u \cos( \alpha ) } })^{2} \\ y = \tan( \alpha ) x + ( \frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} }) {x}^{2} \\ this \: is \: a \: second \: degree \: equation \\ so \: its \: path \: is \: parabola$

(B) 1 Time of flight (t)

$in \: this \: tme \: the \: \\ vertical \: displacemet \: is \: 0 \\ y = 0 \\ u \sin( \alpha ) t - \frac{1}{2} g {t}^{2} = 0 \\ u \: \sin( \alpha ) t \: = \frac{1}{2} g {t}^{2} \\ t = \frac{2u \sin( \alpha ) }{g}$

2 Maximum height (h)

$at \: the \: maximum \: height \: the \: final \: \\ vertical \: component \: of \: velocity \: is \: 0 \\ {v}^{2} - {u}^{2} = 2as \\ 0 - {(u \sin( \alpha ) )}^{2} = - 2gh \\ h = \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g}$

3 Range(r)

$range \: is \: horizontal \: displacement \: \\ during \: time \: of \: flight \\ r = u \cos( \alpha ) t \\ r = u \cos( \alpha ) (\frac{ 2u\sin( \alpha ) }{g} ) \\ r = \frac{ {u}^{2}(2 \sin( \alpha ) \cos( \alpha )) }{g} \\ r = \frac{ {u}^{2} \sin(2 \alpha ) }{g}$