# What is the path of a fragments as they fall to the ground?Why do the fragments follow this path?

2
by nita456

2015-04-08T13:11:21+05:30

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The path of the fragments would depend on the velocity they have.
- If they have 'only vertical component of velocity', then the path would be a straight line. It will travel vertically downward.
- If it has either 'only horizontal component of velocity' or 'both horizontal and vertical component of velocity', its path will be parabolic.
2015-04-12T00:04:58+05:30

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Normally the path of fragments or particles that are projected in two dimensions under the influence of the Earth's gravitational field is a parabola.

Since, we are referring to fragments of a body, it means that a body probably splits into multiple fragments while it is traveling upwards or downwards or when it is on the top of its flight in a vertical line.

At that point of time, the body has only momentum in the vertical direction.  As the momentum is conserved (in the horizontal direction), the fragments split in different horizontal directions, will have velocities such that the total momentum horizontally is zero.  The momentum is conserved here because there is no force acting in the horizontal direction.  In the vertical direction the momentum is not conserved.

Each fragment having some velocity horizontally travels at a uniform speed in the horizontal direction.  The fragment also falls down with the acceleration due to gravity.

The locus of a particle having an acceleration in one direction and a uniform direction in the other moves in a parabolic path.
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derivation of the locus of the path:

let the initial velocity of the fragment of body:  u
Let the initial velocity make an angle of Ф with the horizontal.

the horizontal component of the velocity = u CosФ
This component remains constant as there is no acceleration or deceleration in the horizontal direction.
the vertical component of the velocity = u SinФ
The vertical component changes continuously as Earth's gravitational attraction acts on the fragments.
Hence,  v = u SinФ - g t      at time t

horizontal position of the fragment at time t,  x = u cosФ * t
=> t = x /(u cosФ)        ---- (2)

vertical position of the fragment at time t,  y = u SinФ * t - 1/2 * g * t²    --- (1)
y = u SinФ * x/(u CosФ)  - 1/2 * g * x²/(u CosФ)²              --- by (2)

y = x tan Ф -  g x² / (2u² Cos²Ф)        --- (3)