8 julio, 2024

Oblique parabolic throw: characteristics, formulas, equations, examples

The oblique parabolic shot It is a particular case of the movement of free fall in which the initial velocity of the projectile forms a certain angle with the horizontal, resulting in a parabolic trajectory.

Free fall is a case of movement with constant acceleration, in which the acceleration is that of gravity, which always points vertically downwards and has a magnitude of 9.8 m/s^2. It does not depend on the mass of the projectile, as Galileo Galilei demonstrated in 1604.

If the initial velocity of the projectile is vertical, the free fall has a straight and vertical trajectory, but if the initial velocity is oblique then the free fall trajectory is a parabolic curve, a fact also demonstrated by Galileo.

Examples of parabolic motion are the path followed by a baseball, the bullet fired from a cannon, and the stream of water coming out of a hose.

Figure 1 shows an oblique parabolic shot of 10 m/s with an angle of 60º. The scale is in meters and the successive positions of P are taken with a difference of 0.1 s starting from the initial instant 0 seconds.



The motion of a particle is fully described if its position, velocity, and acceleration as a function of time are known.

The parabolic movement resulting from an oblique shot is the superposition of a horizontal movement at constant velocity, plus a vertical movement with constant acceleration equal to the acceleration of gravity.

The formulas that apply to the oblique parabolic throw are the one that corresponds to a movement with constant acceleration a = gnote that bold type has been used to indicate that acceleration is a vector quantity.

position and velocity

In a movement with constant acceleration, the position depends mathematically on the time in quadratic form.

If we denote r


The bold in the previous expression indicates that it is a vector equation.

Velocity as a function of time is obtained by taking the derivative with respect to t of the position and the result is:


And to obtain acceleration as a function of time, take the derivative of velocity with respect to you resulting:


When time is not available, there is a relationship between velocity and position, which is given by:

v2 =veither2 – 2 g (and – i )


Next we will find the equations that apply to an oblique parabolic throw in Cartesian form.

The movement begins instantly t=0 with starting position (xo, me) and velocity of magnitude veither and angle θthat is to say that the initial velocity vector is ( veither cosθ, veither sinθ ). The movement is accelerating

g = (0, -g).

parametric equations

If the vector formula that gives the position as a function of time is applied and components are grouped and equalized, then the equations that give the coordinates of the position at any instant of time t will be obtained.



Similarly, we have the equations for the velocity components as a function of time.



Where: vox = vo cosθ ; voy =veither sinθ

trajectory equation

y = Ax^2 + Bx + C

A = -g/(2 vox^2)

B = ( voy/vox +gxeither/vox^2)

C = (andeither –voy xeither /vox)


Example 1

Answer the following questions:

a) Why is the effect of air friction usually neglected in parabolic draft problems?

b) Does the shape of the object have any importance in the parabolic shot?


a) For the movement of a projectile to be parabolic, it is important that the force of air friction be much less than the weight of the object that is thrown.

If a ball made of cork or some light material is thrown, the friction force is comparable to the weight and its trajectory cannot approach a parabola.

On the contrary, if it is a heavy object like a stone, the friction force is negligible compared to the weight of the stone and its trajectory does approach a parabola.

b) The shape of the thrown object is also relevant. If a sheet of paper is thrown in the shape of a small plane, its movement will not be free fall or parabolic, since the shape favors air resistance.

On the other hand, if the same sheet of paper is compacted into a ball, the resulting movement is very similar to a parabola.

Example 2

A projectile is launched from the horizontal ground with a speed of 10 m/s and an angle of 60º. These are the same data with which Figure 1 was made. With these data, find:

a) Instant at which it reaches the maximum height.

b) The maximum height.

c) The speed at the maximum height.

d) The position and velocity at 1.6 s.

e) The instant at which it hits the ground again.

f) The horizontal reach.

Solution to)

The vertical velocity as a function of time is


At the moment the maximum height is reached, the vertical velocity is zero for an instant.

8.66 – 9.8 t = 0 ⇒ t = 0.88 s.

Solution b)

The maximum height is given by the coordinate and for the instant at which that height is reached:

y(0.88s) = i + go t -½ gt^2 = 0 + 8.66*0.88-½ 9.8 0.88^2 =


Therefore the maximum height is 3.83 m.

Solution c)

The velocity at maximum height is horizontal:


Solution d)

The position at 1.6 s is:

x(1.6) = 5*1.6 = 8.0m

y(1.6) = 8.66*1.6-½ 9.8 1.62 = 1.31m

Solution e)

When the coordinate touches the ground and is cancelled, then:


Solution f)

The horizontal range is the x coordinate just at the instant it hits the ground:

x(1.77) = 5*1.77 = 8.85m

Example 3

Find the equation of the trajectory with the data from Example 2.


The parametric equation of the trajectory is:



And the Cartesian equation is obtained by isolating t from the first and substituting in the second

y = 8.66*(x/5)-½ 9.8 (x/5)^2


y = 1.73x – 0.20x^2


P.P. Teodorescu (2007). «Kinematics». Mechanical Systems, Classical Models: Particle Mechanics. Springer.
Resnick, Halliday, & Krane (2002). Physics Volume 1. Cecsa, Mexico.
Thomas Wallace Wright (1896). Elements of Mechanics Including Kinematics, Kinetics and Statics. E and FN Spon.
Wikipedia. Parabolic movement. Retrieved from es.wikipedia.org.
Wikipedia. Projectile motion. Retrieved from en.wikipedia.org.

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