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.
formulas
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) = r o + v 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) = v 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:
v 2 = v or 2 – 2 g (y – yo)
Equations
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 v 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.
v x (t) = v ox
v e (t) = v o – gt
Where: v ox = v or cosθ; v oy = v o sinθ
path equation
y = A x ^ 2 + B x + C
A = -g / (2 v ox ^ 2)
B = (v oy / v ox + gx o / v ox ^ 2)
C = (y o – v oy x o / v ox )
Examples
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?
Answers
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
v 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:
v 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 2 = 1.31 m
Solution e)
When you touch the coordinate floor and it cancels, then:
y (t) = 8.66 * t-½ 9.8 t 2 = 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.
Solution
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
