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Examples of Parabolic motion

parabolic movement or parabolic motion is called the displacement of an object whose trajectory traces the shape of a parabola.

The parabolic movement is characteristic of an object or projectile subject to the laws of a uniform gravitational field that crosses a medium of little or no resistance to advancement , and is considered the conjunction of two movements different from the simultaneous: a uniform horizontal displacement and another vertical accelerated .

To tell the truth, this type of movement, so usual in objects on the earth’s surface, points rather to the elliptical: any object that we throw will try to trace an ellipse with one of its focuses in the gravitational center of our planet . It will not succeed, it will hit the ground and stop its movement, but its trajectory will be that of an ellipse segment, coinciding in turn with a parabola .

For this reason the formulas of the parabola are used to calculate this type of movement: their equations are much simpler than those of the elliptical movement.

In addition, the parabolic motion always obeys the following considerations :

  • Knowing the initial angle of inclination, the initial velocity and the height difference between the departure point and the arrival point, the entire trajectory can be calculated.
  • The departure and arrival angles are always identical, unless the height of both points varies.
  • With a 45 ° angle, the greatest distance covered is achieved.
  • Given a fixed angle, the distance can be increased by increasing the speed.
  • The vertical movement can be analyzed without taking the horizontal into account.

Examples of parabolic movement

  1. The firing of a military projectile (artillery charge, mortar, etc.) from the barrel of the cannon to the point of fall or target.
  2. The chute of a soccer ball from the archery to fall in the opposite field.
  3. The trajectory of a golf ball during the initial long distance shot.
  4. The water jet from a hose like those used by firefighters to put out a fire.
  5. The water jet of rotating sprinklers in a garden or park, throwing the liquid around with a uniform speed and angle.
  6. The throwing of a stone when we try to knock down fruits from a tree but we miss them and fall from the other side.
  7. A volleyball serve that makes the ball rise above the net and fall with the same angle of inclination on the other side.
  8. The launching of a bomb or missile from a plane in full flight, is a semi-parabolic movement because it covers half of a parabola (but responds to the same physical considerations).
  9. The launching of a disc like those that are thrown to practice the target shooting with a rifle.
  10. The rebound of a stone on the surface of the water will draw small parabolas smaller and smaller with each bounce, until it loses the initial thrust and sinks.

Examples of parabolic motion exercises

Someone kicks a soccer ball and it is thrown at an angle of 37 ° and with a speed of 20 m / s. Knowing that the gravitational constant is 9.8 m / s ^ 2, calculate: a) the maximum height of the ball, b) the total time it remains in the air, c) the distance it has traveled when it fell.

Resolution :

1.

Vox = Vo Cos a = 20 m / s Cos 37 ° = 15.97 m / s

Voy = Vo Sen a = 20 m / s Sen 37 ° = 12.03 m / s

2. To obtain the maximum height time:

Voy = 0

Therefore: t = (Vfy – Voy) / g = (0 – 12.03 m / s) / 9.8 = 1.22 s

3. To obtain the maximum height:

Ymax = t + gt going 2 /2 = 12.03 m / s (1.22 s) + ((-9.8m / s 2 ) (1.22 s) 2 ) / 2 = 7.38 m

4. To obtain the total time, simply multiply the time of maximum height by 2, since we know that the trajectory in this case is symmetrical: the projectile will take twice as long to fall as it did to reach its maximum height.

Total T = tmax (2) = 1.22s (2) = 2.44 s

5. To obtain the maximum scope the formula will be used:

X = Vx total t = 15.97 m / s (2.44 s) = 38.96 m

6

Vfy = gt + Voy = (- 9.8) (1 s) + 12.03 m / s = 2.23 m / s

Vfx = 15.97 m / s since it is constant throughout the movement.

 

An involuntary artillery shot occurs at a speed of 30 m / s, forming an angle of 60 ° to the horizon. To alert the civilian population, it is necessary to calculate the total distance covered, the maximum height and the time of fall of the shot.

Resolution :

  • To obtain the distance traveled:

d = v 2 sin 2 a / g = (30m / s) 2 sin 2 (60 °) / 9.8 m / s 2 = 158.99 m

  • To obtain the height reached:

h = v 2 sin 2 a / 2g = (30 m / s) 2 sin 2 (60 °) / 2 (9.8 m / s 2 ) = 36.29 m

  • To obtain the total time:

t = v 1 sin a / g = 30 m / s (sin 60 °) / 9.8 m / s 2 = 2.85 s

 

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