Oblique parabolic shot characteristics, formulas, equations, examples

The oblique parabolic shot is a particular case of free fall motion 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 motion with constant acceleration, in which 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 mass of the projectile, as Galileo Galilei showed 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 trajectory of free fall is a parabolic curve, a fact also demonstrated by Galileo.

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

Figure 1 shows an oblique parabolic draft 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.

Article index

• 1 Formulas
  • 1.1 Position and speed 
• 2 Equations
  • 2.1 Parametric equations
  • 2.2 Equation of trajectory
• 3 Examples 
  • 3.1 Example 1
  • 3.2 Example 2
  • 3.3 Example 3
• 4 References

Formulas

The motion of a particle is fully described if its position, its velocity, and its 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 speed, plus a vertical motion with constant acceleration equal to the acceleration of gravity..

The formulas that are applied to the oblique parabolic shot are those that correspond to a movement with constant acceleration a = g, note that bold has been used to indicate that the acceleration is a vector quantity.

Position and speed 

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

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

r(t) = r_or + v_or t + ½ g t^two

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

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

v(t) = v_or + g t

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

to(t) = g

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

v^two = vor^two - 2 g (and - me)

Equations

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

The movement begins in the instant t = 0 with starting position (xo, me) and speed of magnitude vor and angle θ, that is, the initial velocity vector is (vor cosθ, vor senθ). The movement proceeds with acceleration 

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.

x (t) = x_or + v_ox t 

y (t) = y_or + v_hey t -½ g t^two

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

v_x(t) = v_ox

v_Y(t) = v_hey - g t 

Where: v_ox = vor cosθ; v_hey = vor senθ

Equation of the trajectory

y = A x ^ 2 + B x + C

A = -g / (2 vox^ 2)

B = (vhey/ vox + g xor/ vox^ 2)

C = (andor - vhey xor / vox)

Examples 

Example 1

Answer the following questions:

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

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

Answers

a) For the movement of a projectile to be parabolic, it is important that the friction force of the air is much less than the weight of the object being 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 such as 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 an airplane, 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 elaborated. With these data, find:

a) Moment in which it reaches the maximum height.

b) The maximum height.

c) Speed ​​at maximum height.

d) Position and velocity at 1.6 s.

e) The moment it hits the ground again.

f) The horizontal reach.

Solution to)

The vertical speed as a function of time is

v_Y(t) = v_hey - g t = v_or sinθ - g t = 10 sin60º - 9.8 t = 8.66 - 9.8 t 

At the moment the maximum height is reached, the vertical speed 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 Y for the instant in which that height is reached:

and (0.88s) = I + go t -½ g t ^two = 0 + 8.66 * 0.88-½ 9.8 0.88 ^two = 

3.83 m

Therefore the maximum height is 3.83 m.

Solution c)

The speed at maximum height is horizontal:

v_x(t) = v_ox = v_or cosθ = 10 cos60º = 5 m / s 

Solution d)

The position at 1.6 s is:

x (1.6) = 5 * 1.6 = 8.0 m

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

Solution e)

When the y coordinate touches the ground, then:

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

Solution f)

The horizontal reach is the x coordinate just at the instant it touches the ground:

x (1.77) = 5 * 1.77 = 8.85 m

Example 3

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

Solution 

The parametric equation of the path is:

x (t) = 5 * t

y (t) = 8.66 * t-½ 9.8 t ^two

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

y = 8.66 * (x / 5) -½ 9.8 (x / 5) ^two

Simplifying:

y = 1.73 x - 0.20 x ^ 2 

References

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