Oblique Parabolic Shot: Characteristics, Formulas, Equations, Examples

The oblique parabolic shot is a particular case of free fall movement in which the projectile’s initial velocity is at a certain angle to the horizontal, resulting in a parabolic path.

Free fall is a case of motion with constant acceleration, where the acceleration is that of gravity, which always points vertically downward and has a magnitude of 9.8 m/s^2. It does not depend on the projectile’s mass, like 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, the path of the free fall is a parabolic curve, a fact also demonstrated by Galileo.

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


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

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

The formulas that apply to the oblique parabolic shot are those that correspond to a motion with constant acceleration a = g , note that bold was used to indicate that the acceleration is a vector quantity.

position and speed

In a motion with constant acceleration, position is mathematically dependent on time in quadratic form.

If we denote r (t) the position at time t , r or the position at the initial instant, v or the initial velocity, g the acceleration and t = 0 as the initial instant, the formula that gives the position for each moment of time t is :

r (t) = o + or t + ½ g t 2

The bold letters in the above expression indicate that it is a vector equation.

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

v (t) = o + g t

And the function for accelerating the time is taken the derivative of velocity with respect to T, resulting in:

a (t) = g

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

2 = v or 2 – 2 g (y – yo)


Below, we will find the equations that apply to an oblique parabolic shot in Cartesian form.

The movement starts at time t = 0 with the initial position (xo, i) and velocity of magnitude o and angle θ , that is, the initial velocity vector is (v or cosθ, v or sinθ) . The movement takes place with acceleration

g = (0, -g).

parametric equations

If the vector formula that gives the position as a function of time is applied and the components are grouped and combined, the equations that give the coordinates of the position at any point in time are obtained.

x (t) = x o + v ox t

y (t) = y o + v oy t -½ gt 2

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

x (t) = v ox

e (t) = v o – gt

Where: ox = v or cosθ; oy = v sinθ

path equation

y = A x ^ 2 + B x + C

A = -g / (2 v ox ^ 2)

B = (v oy / v ox + gx / v ox ^ 2)

C = (y – v oy / v ox )


Example 1

Answer the following questions:

a) Why, in parabolic shooting problems, is the effect of friction with air generally neglected?

b) Is the shape of the object in the parabolic firing important?


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

If a ball of cork or some light material is thrown, the frictional 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 frictional force is negligible compared to the weight of the stone and its trajectory approximates to a parabola.

b) The shape of the launched object is also relevant. If a sheet of paper in the shape of an airplane is thrown, its movement will not be free-fall or parabolic, as the shape favors air resistance.

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

Example 2

A projectile is launched from the horizontal ground with a velocity of 10 m/s and an angle of 60°. These are the same data with which Figure 1. It was elaborated with the following data:

a) Instant when reaching maximum height.

b) The maximum height.

c) Speed ​​at maximum height.

d) Position and velocity in 1.6 s.

e) The moment he touches the ground again.

f) The horizontal reach.

Solution a)

Vertical velocity as a function of time is

y (t) = v oy – gt = v or sinθ – gt = 10 sen60º – 9.8 t = 8.66 – 9.8 t

When 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 when that height is reached:

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

3.83 m

Therefore, the maximum height is 3.83 m.

Solution c)

The maximum height velocity is horizontal:

x (t) = v ox = v or cosθ = 10 cos60º = 5 m / s

Solution d)

The position in 1.6 s is:

x (1.6) = 5 * 1.6 = 8.0 m

and (1.6) = 8.66 * 1.6-½ 9.8 1.6 = 1.31 m

Solution e)

When you touch the coordinate floor and it cancels, then:

y (t) = 8.66 * t-½ 9.8 t = 0 ⇒ t = 1.77 s

Solution f)

The horizontal band is the x coordinate at the moment it touches the ground:

x (1.77) = 5 * 1.77 = 8.85 m

Example 3

Find the path equation with the data in example 2.


The parametric path equation is:

x (t) = 5 * t

y (t) = 8.66 * t-½ 9.8 t ^ 2

And the Cartesian equation is obtained by clearing t from the first and substituting for the second

y = 8.66 * (x / 5) -½ 9.8 (x / 5) ^ 2

Simply put:

y = 1.73 x – 0.20 x ^ 2

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