Question

Show that the path of a projectile is a parabola

Answer 1

Let a body is projected with speed $um/s$ inclined $\theta$ with horizontal line$.$

Then$,$ vertical component of $u,$ $u_y=u\cos \theta$

Horizontal component of $u_x=u\sin \theta$

acceleration on horizontal$,$ $ax=0$

acceleration on vertical$,$ $ay=-g$

Now$,$ use formula

$X=u_{x}t$

$X=u\cos \theta .t$

$t=X/u\cos \theta$--------------------------$\left(1\right)$

Again$,$ $y=u_{y}t+1/2a_y{t}^2$

$y=u\sin \theta t-1/2g{t}^2$

put equation $\left(1\right)$ here$,$

$y=u\sin \theta \times x/u\cos \theta -1/2g{t}^2\times x^2/u^2{\cos }^2\theta$

$=\tan \theta x-1/2g{x}^2/u^2{\cos }^2$

$y=\tan \theta .x-1/2g{x}^2/u^2{\cos }^2\theta$

this equation is similar to standard equation of parabola $y=a{x}^2+bx+c$ her$,$ $a,b$and $c$ are constant

So$,$ A projectile motion is parabolic motion$.$