Answer:
Lets take
u =velocity
θ =angle from horizontal
The horizontal component of velocity = u cosθ
The vertical component of velocity = u sinθ
The horizontal distance x in time t
x = u cosθ .t -----------1
The vertical distance y in time t
[tex]y=usin\theta.t-\dfrac{1}{2}gt^2[/tex] --------2
t= x/(u cosθ) ( tanθ=sinθ/cosθ)
[tex]y=xtan\theta-\dfrac{gx^2}{2u^2cos^2\theta}[/tex]
Form above we can say that path of the projectile motion is parabolic.
