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It isn't. Any projectile path is elliptical or round. One-half of an ellipse in polar coordinates mapped (not just converted) into Cartesian coordinates is a parabola. If you get this far, any projectile path on a spherical Earth's surface would be elliptical, but if you turned the Earth inside out, the path would be hypocycloidal. Hope this helps. ===========================================

??? An ORBIT is elliptical. But the behavior of an ordinary projectile over non-planetary distances is effectively that of an object launched at an angle with regard to a plane in a uniform gravitational field - over short distances the earth's surface is reasonably close to a plane and gravity acts essentially perpendicular to that plane. So, an object launched at an angle θ with a velocity of vo moves in two dimensions. It has a constant velocity vo * cos(θ) horizontally and a downward acceleration that's proportional to sin(θ) and the acceleration of gravity g.

The constant horizontal velocity means that the horizontal (x-axis) distance covered is directly proportional to the flight time, while the accelerating vertical velocity means the vertical (y-axis) distance is proportional to the square of the flight time. But that's exactly the relationship that describes a parabola; i.e. y = kx2

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16y ago
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11y ago

The projectile doesn't act as a parabola, but a graph of the magnitude of its

velocity vs time does.

In projectile motion, the horizontal component of velocity is constant and the

vertical component is accelerated. The 2-dimensional resultant of one constant

component and an accelerated one is a parabola.

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14y ago

It will only approximate a parabolic path (you have to ignore the effect of air resitance, the coriolis effect, the reduction of gravity at higher altitudes etc.) If we do ignore all these (rater small) effects the simple answer is that there is a gravitational field that operates on the projectile. The initial velocity of the projectile is acted upon by the vector of gravity.

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12y ago

Parabolic Path of a Projectile

Easily:

In the vertical direction

F=ma=md2ydt2=−mg

Integrated twice we obtain

y(t)=y(0)+vy(0)t−12gt2

Which is indeed the equation of a parabola with respect to time.

Now in the x-direction, we have

x(t)=x(0)+vx(0)t

since there is no force in this direction. It should be clear that if we re-write y(t) in terms of x(t), by solving for t in the above equation, y(t) will also be a parabola in terms of x.

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8y ago

It is, to a first approximation, assuming air resistance can be neglected, and assuming the path is relatively short. Actually, any object moving around Earth travels in an ellipse; so the parabola is only an approximation - valid if you consider Earth's surface to be flat, and gravity not to change with altitude.

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14y ago

The gravitational attraction on the projectile will pull it pack down to Earth.

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8y ago

The path of a projectile is a parabola. This is a two dimensional motion.

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15y ago

Its trajectory.

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8y ago

Yes, that is usually correct.

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13y ago

trajectory

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Q: Is the shape of a path followed by a projectile a parabola?
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