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the motion of a projectile

Galileo was the first in a scientific way to study the motion of a projectile showing that its trajectory is a parabola. The results are published in the work "Discourses and mathematical demonstrations concerning two new sciences."

Obtain the results of Galileo by the equations of motion, taking into account only the gravitational forces acting on the projectile, considered as a material point, and neglecting air friction.

We choose a reference system with the y-axis positive upward, so that the origin of the axes is the point (0 x, y 0) = (0,0) of departure of the projectile; The components will be x = 0, y = - g.

Using the law of fall of a serious, we draw the trajectory of a bullet, making sure it is a parable and then showing some more features.

The velocity vector v at the initial instant t = 0 has form v 0 and is tilted at an angle θ with respect to the positive direction of x, its components are:

v 0x = v 0 cos θ

_{0Y v = v 0} senθ

The law of motion, which expresses the speed is a function of time t (v (t) = v 0 + at).
Since there are no horizontal components of acceleration, the horizontal component of velocity v x remains constant, the vertical component v y change over time because there is a constant downward acceleration (a y = - g):

v x = v 0x

v y = v 0Y - gt

The velocity vector is tangent to the trajectory at each point, its form is not constant and can be obtained by applying the Pythagorean theorem.

The laws of motion describing the motion of the projectile in space are those of a rectilinear motion along x and y uniformly accelerated along, independent of each other. Then the coordinates of the projectile in parametric form (the parameter is the time t) at a generic instant t are:

x (t) = v 0x

t y (t) = v 0Y t - 1 / 2GT ^ 2

From this equation it is possible to obtain the equation of the trajectory in Cartesian form, obtaining t from the first equation and substituting in the second. You get the equation of the trajectory of the bullet:

as you can see that is a downward parabola through the origin of the axes. A drawn representation of motion with the velocity components is shown below.

The vertex of the parabola can be found mathematically by the known relationship V = (-b/2a; -Δ/4a). Arguing from a physical point of view, the vertex of the parabola is obtained by requiring that the velocity along y is 0. It is then the point:

M x represents the abscissa of the point maximum height, y M the maximum height reached by the projectile.

To calculate the range, that is the point at which the bullet falls on the axis of x, y sufficient to impose (x) = 0, ie to the intersection of the parabolic trajectory of the bullet with the x-axis You get two solutions:

The first is obviously the source, the second is the range sought.

The time taken to travel x G is called time of flight (t Flight ) and coincides with twice the time necessary to reach the maximum height y M and return to the soil

t Flight G = x / v = 2x 0x M / v 0x .

Note that the location where the bullet touches the x-axis, the speed is the same in the form at the start but it is symmetrical with respect to x.

The launch angle for maximum range is where you can get as follows:

sen2θ = 1 → 2θ = 90 ° → θ = 45 °.