Question

show that path of projectile motion is parabolic

Answer 1

Let's consider an object is projected with initial velocity (u) making angle of projection with respect to ground moving freely during the motion under the action of gravity.. so time is given by

x = horizontal velocity × time
$x = u \cos( \alpha ) \times t$
therefore time is given by
$t = \frac{x}{u \cos( \alpha ) }$
.....(1)

now the path traced by an object during its motion is given by 2nd kinematical equation

$s = ut + \frac{1}{2} a {t}^{2}$
then with respect to y-axis we get
substitute s=0, u = usin, a= -g

$y= (u \sin( \alpha ) )t - \frac{1}{2} g {t}^{2}$
then substitute equation 1 in above equation we get

$y = (u \sin( \alpha ) )( \frac{x}{u \cos( \alpha ) } ) - \frac{1}{2} g ({ \frac{x}{u \cos( \alpha ) } })^{2}$
after simplification we get
$y = \tan( \alpha ) x - ( \frac{g}{2 {u}^{2} { \cos( \alpha ) }^{2} } ) {x}^{2}$
this equation is called trajectory i.e path of projectile
as angle, gravity and velocity is constant throughout the motion then this path can be given as
as it is parabolic in nature

$y = \alpha x - \beta {x}^{2}$
this equation is of parabola