# 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