Question

Show that the path of a projectile is a parabola.

Answer 1

Parabolic path of projectile:

Let a body is projected with speed $u\mathrm{m}/\mathrm{s}$ inclined $\theta$ with horizontal line .

Then, vertical component of, $u=u\sin \theta$

Horizontal component of, $u=u\cos \theta$

Acceleration on horizontal, $ax=0$

Acceleration on vertical, $ay=-g$

Now, use formula,

$$\begin{gathered} x=u\cos \theta \times t \\ \therefore t=\frac{x}{u\cos \theta }...\left(1\right) \end{gathered}$$

Again, $y=u\sin \theta t-\frac{1}{2}g{t}^2$

Put equation $\left(1\right)$

$$\begin{gathered} y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{1}{2}g\times \frac{x^2}{u^2{\cos }^2\theta } \\ =\tan \theta x-\frac{g{x}^2}{2u^2{\cos }^2\theta } \end{gathered}$$

This equation is similar to Standard equation of parabola $y=a{x}^2+bx+c$ here, $a,b$ and $c$ are constant .

Hence, a projectile motion is a parabolic motion.