1.

nat isa Protectile? Derive anecile

Answer»

Case1: Projectile projectected parallel to horizontal :

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

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

Putting value oftfrom (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 oftffrom (1) into (2)

We get Y = X tanβ - g X²/2u²ccos²β

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


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