Question

what is projectile derive the expression of vertical height and horizontal range ​

Answer 1

Explanation:

vertical height

${v}^{2} = {u}^{2} + 2as \\ 0 = (usin \alpha ) ^{2} + 2( - g)h \\ 2gh = {u}^{2} { \sin^{2} ( \alpha ) } \\ h = \frac{ {u}^{2} { \sin^{2} ( \alpha ) }}{2g}$

horizontal range

$s = ut + \frac{1}{2}a {t}^{2} \\ r = u \cos( \alpha ) \times \frac{2u \sin( \alpha ) }{g} + \frac{1}{2} \times 0 \times t \\ r = \frac{ {u}^{2} 2 \sin( \alpha ) \cos( \alpha ) }{g}$