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

Answer 1

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

Answer 2

Yes, the shape of the path followed by a projectile in a uniform gravitational field is a parabola. This is due to the fact that the motion of the projectile can be broken down into independent horizontal and vertical components, resulting in a parabolic trajectory.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

Its trajectory.

Answer 10

Yes, that is usually correct.