Question

Derive an equation of the path of a projectile.

Answer 1

.an equation of the path of a projectile.

Answer 2

Case 1

Projectile projected parallel to horizontal :

Motion along x axis: x= u t ...... (1)

Motion along y axis, y = g t²/2 ......(2)

Putting value of t from (1) into (2)

We get, y = g x²/2u² ......(3)

This represents a parabola.

Case 2

When projectile is given angular projection:

Let the projectile makes angle β with horizontal and is projected with velocity u.

Ux = u cos β and Uy =u sinβ

Ax (acceleration along x axis) = 0

Ay = - g (acceleration along y axis)

Now, motion along x axis, X = u cosβ t ......(1)

And motion along y axis, Y = u sinβ t - gt²/2 ...(2)

Putting value of t ffrom (1) into (2)

We get Y = X tanβ - g X²/2u² ccos²β

This represents a parabola. Hence the projectile follows the parabolic path.