Question

What is "Trajectory of a projectile"?

Answer 1

Assume a projectile of mass $m$ is projected with velocity $u$ at an angle $\theta$ with the ground.

The horizontal component of the initial velocity is given as,

$u_x=u\cos \theta$

The horizontal displacement at any time $t$ is given as,

$X=u_{x}t$

$t=\frac{X}{u\cos \theta }$

The vertical component of the initial velocity is given as,

$u_y=u\sin \theta$

The horizontal displacement at any time $t$ is given as,

$Y=u_{y}t-\frac{1}{2}g{t}^2$

$Y=u\sin \theta \times \frac{X}{u\cos \theta }-\frac{1}{2}g{\left(\frac{X}{u\cos \theta }\right)}^2$

$Y=X\tan \theta -\frac{1}{2}g{\left(\frac{X}{u\cos \theta }\right)}^2$

The above equation is a parabola.

Thus, the path of a projectile is parabolic.