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

Suppose the object is launched with angle α in relation to the horizontal and initial velocity 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:

ox = v or cos α

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:

Related:   Principle of Transmissibility of Force (Resolved Exercises)

-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 y = 0 . Knowing that there is the following relationship between position and the square of velocity:

2 = v oy 2 – 2gy

Making y = 0 once the maximum height is reached:

 0 = v oy 2 – 2g E max → y max  = v oy 2 / 2g

With:

oy = v or sinα

maximum time

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

y = v or .sen α  – gt

Knowing that y becomes 0, when t = t max , results:

o .sen α  – gt max = 0

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 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 final position :

and final = v or .sen α .t v – ½g.t 2

Anyway, the maximum horizontal range is:

Related:   Kepler’s Laws: explanation, exercises, experiment

max = v ox . you see

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

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:

max = v ox . t flight = v o .cos α . you see

When you land later, the height will return to 0, so:

the . sin α. v – ½ gt 2 = 0

Since v is a common factor, it is simplified:

the . sin α  – ½g.t v = 0

We can clear t v from the first equation:

v = x max / v o .cos α

And replace in the second:

the . sin α  – (½ gx max / v or .cos α ) = 0

By multiplying all terms by or .cos α,  the expression is not changed and the denominator disappears: 

(v o . sin α.) (v o .cos α ) – ½g.x max = 0

or 2 sin α. cos α  = ½g.x max

Now you can clean v or or also replace the following identity:

Related:   Wavelength: characteristics, formulas and exercise

sin 2α = 2 sin α. cos α  → v or 2 sin 2α = gx max

It is calculated or 2 :

or 2 = g. max / sin 2α = (9.8 x 0.8 / sin 110) m 2 / s 2 = 8.34 m 2 / s 2

And finally the maximum height:

 max = v 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:

 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:

max = v or 2 sin 2a / g

Changing the angle also changes the horizontal range:

 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.

Related Articles

Leave a Reply

Your email address will not be published.

Back to top button