# Parabolic Shot: Characteristics, Formulas and Equations, Examples

The **parabolic** of launching an object or projectile angle and letting it move under gravity. If air resistance is not considered, the object, regardless of its nature, will follow a parabola arc path.

It is an everyday movement, as among the most popular sports are those in which balls or balls are thrown, either with the hand, foot or with an instrument such as a racket or bat, for example.

For your study, the parabolic shot is divided into two superimposed movements: a horizontal one without acceleration and a vertical one with constant acceleration, which is gravity. Both movements have initial speed.

Let’s say the horizontal movement occurs along the x axis and vertical movement along the y axis. Each of these movements is independent of the other.

As determining the projectile’s position is the main objective, it is necessary to choose an appropriate reference system. Details are below.

__Parabolic Throw Formulas and Equations__

__Parabolic Throw Formulas and Equations__

Suppose the object is launched with angle α in relation to the horizontal and initial velocity **v **_{or} as shown in the figure below on the left. The parabolic throw is a movement that occurs in the *xy* plane and, in this case, the initial velocity is reduced like this:

*v _{ox} = v _{or} cos α*

*v _{oy} = v _{o} sin α*

The projectile’s position, which is the red dot in Figure 2, right image, also has two time-dependent components, one at *x* and one at *y* . Position is a vector denoted by **r and** its units are length.

In the figure, the starting position of the projectile coincides with the origin of the coordinate system, so x _{o} = 0, y _{o} = 0. It is not always the case, you can choose the origin anywhere, but this choice greatly simplifies calculations.

As for the two movements in x and y, they are:

-x (t): is a straight, uniform motion.

-y (t): corresponds to a uniformly accelerated rectilinear movement with g = 9.8 m / s ^{2} and pointing vertically downwards.

In mathematical form:

*x (t) = v _{or} cos α *

*.t*

*y (t) = v _{or} .sin α *

*.t – ½g.t*

^{2}The position vector is:

*r** (t) = [v _{o} cos α *

*.t]*

**i**+ [v_{o}.sen α*.t – ½g.t*

^{2}]**j**In these equations, the attentive reader will notice that the minus sign is due to the fact that gravity points to the ground, the direction chosen as negative, while the ascent is considered positive.

As velocity is the first derivative of position, just derive **r** (t) with respect to time and get:

*v** (t) = v _{o} cos α *

**i +**(v_{o.}sin α*– gt)*

**j**Finally, acceleration is expressed vectorally as:

* **a** (t) = -g j*

**– Trajectory, maximum height, maximum time and horizontal reach**

**Trajectory**

To find the explicit equation of the path, which is the curve y(x), the time parameter must be eliminated, solving the equation for x(t) and replacing y(t). Simplification is quite a bit of work, but you finally get:

**Maximum height**

The maximum height occurs when *v _{y} = 0* . Knowing that there is the following relationship between position and the square of velocity:

*v _{y }^{2} = v _{oy }^{2} – 2gy*

Making *v _{y} = 0* once the maximum height is reached:

* **0 = v _{oy }^{2} – 2g E _{max} → y _{max } = v _{oy }^{2} / 2g*

With:

*v _{oy} = v _{or} sinα*

**maximum time**

The max time is the time it takes the object to arrive and _{max} . To calculate, we use:

*v _{y} = v _{or} .sen α *

*– gt*

Knowing that *v _{y}* becomes 0, when

*t = t*, results:

_{max}*v _{o} .sen α *

*– gt*

_{max}= 0*t _{max} = v _{oy} / g*

**Maximum horizontal range and flight time**

Reach is very important as it signals where the object will land. That way we’ll know if it hits the target or not. To find it, we need flight time, total time or _{v} .

From the previous illustration, it is easy to conclude that *t _{v} = 2.t _{max}* . But be carefulǃ this is only true if the launch is level, ie the height of the start point is equal to the height of the finish. Otherwise, time is found by solving the quadratic equation resulting from the substitution of the

_{final}*and*position :

_{final}*and _{final} = v _{or} .sen α *

*.t*

_{v}– ½g.t_{v }^{2}Anyway, the maximum horizontal range is:

*x _{max} = v _{ox} . you _{see}*

__Parabolic Shot Examples__

__Parabolic Shot Examples__

Parabolic shooting is part of the movement of people and animals. Also from almost every sport and game in which gravity intervenes. For example:

**Parabolic Shot in Human Activities**

-The stone thrown by a catapult.

-The goalkeeper’s kick.

-The ball played by the pitcher.

-The arrow coming out of the bow.

-All types of jumps

-Throw a stone with a sling.

-Any throwing weapon.

**The parabolic shot in nature**

-Water that comes from natural or artificial jets, such as those from a fountain.

Stones and lava sprouting from a volcano.

-A ball that bounces on the sidewalk or a rock that bounces in water.

-All kinds of jumping animals: kangaroos, dolphins, gazelles, cats, frogs, rabbits or insects, to name a few.

__Exercise__

__Exercise__

A grasshopper jumps at an angle of 55 degrees from the horizontal and drops 0.80 meters later. Meet:

a) The maximum height reached.

b) If he jumped at the same initial speed but at a 45° angle, would he reach higher?

c) What can be said about the maximum horizontal range for this angle?

**Solution for**

When the data provided by the problem does not contain the initial velocity v _{or} the calculations are a little more laborious, but a new expression can be derived from the known equations. Starting from:

*x _{max} = v _{ox} . t _{flight} = v _{o} .cos α *

*. you*

_{see}When you land later, the height will return to 0, so:

*v _{the} . *sin α.

*t*

_{v}– ½ gt_{v }^{2}= 0Since *t _{v}* is a common factor, it is simplified:

*v _{the} . *sin α

*– ½g.t*

_{v}= 0We can clear t _{v} from the first equation:

*t _{v} = x _{max} / v _{o} .cos α*

And replace in the second:

*v _{the} . *sin α

*– (½ gx*

_{max}/ v_{or}.cos α*) = 0*

By multiplying all terms by *v _{or} .cos α,*

*the expression is not changed and the denominator disappears:*

*(v _{o} .* sin α.)

*(v*

_{o}.cos α*) – ½g.x*

_{max}= 0*v _{or }^{2}* sin α.

*cos α*

*= ½g.x*

_{max}Now you can clean v _{or} or also replace the following identity:

sin 2α = 2 sin α. *cos α ** → v _{or }^{2}* sin 2α =

*gx*

_{max}It is calculated *v _{or }^{2}* :

*v _{or }^{2} = g. *

*x*sin 2α = (9.8 x 0.8 / sin 110) m

_{max}/^{2}/ s

^{2}= 8.34 m

^{2}/ s

^{2}

And finally the maximum height:

** ***y _{max} = v _{y }^{2} /2 g = (8.34 x sin ^{2} 55) / (2 x 9.8) m = 0.286 m = 28.6 cm*

* ***Solution b**

The lobster manages to maintain the same horizontal velocity, but decreasing the angle:

* **y _{max} = v _{oy }^{2} / 2g = (8.34 x sin ^{2} 45) / (2 x 9.8) m = 0.213 m = 21.3 cm*

Reach a lower height.

**Solution c**

The maximum horizontal range is:

*x _{max} = v _{or }^{2}* sin 2a /

*g*

Changing the angle also changes the horizontal range:

* **x _{max} = 8.34* sin 90 /

*9.8*m = 0.851 m = 85.1 cm

The jump is longer now. The reader can check if the maximum angle is 45º because:

sin 2α = sin 90 = 1.