- Parabolic shot formulas and equations
- - Trajectory, maximum height, maximum time and horizontal reach
- Trajectory
- Maximum height
- Maximum time
- Maximum horizontal reach and flight time
- Examples of parabolic shooting
- Parabolic shooting in human activities
- The parabolic shot in nature
- Exercise
- Solution to
- Solution c
- References
The Parabolic of throwing an object or projectile angle and let it move under the action of gravity. If air resistance is not considered, the object, regardless of its nature, will follow a parabola arc path.
It is a daily movement, since among the most popular sports are those in which balls or balls are thrown, either with the hand, with the foot or with an instrument such as a racket or a bat for example.
Figure 1. The water jet from the ornamental fountain follows a parabolic path. Source: Wikimedia Commons. Zátonyi Sándor (ifj.), Fizped / CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0)
For its study, the parabolic shot is broken down into two superimposed movements: one horizontal without acceleration, and the other vertical with constant downward acceleration, which is gravity. Both movements have initial speed.
Let's say that the horizontal motion runs along the x-axis and the vertical motion along the y-axis. Each of these movements is independent of the other.
Since determining the position of the projectile is the main objective, it is necessary to choose an appropriate reference system. The details follow.
Parabolic shot formulas and equations
Suppose the object is thrown with angle α with respect to the horizontal and initial velocity v or as shown in the figure below left. The parabolic shot is a movement that takes place on the xy plane and in that case the initial velocity is decomposed as follows:
Figure 2. On the left the initial velocity of the projectile and on the right the position at any instant of the launch. Source: Wikimedia Commons. Zátonyi Sándor, (ifj.) Fizped / CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0).
The position of the projectile, which is the red dot in Figure 2, right image, also has two time-dependent components, one at x and the other at y. Position is a vector denoted r and its units are length.
In the figure, the initial position of the projectile coincides with the origin of the coordinate system, therefore x o = 0, and o = 0. This is not always the case, you can choose the origin anywhere, but this choice simplifies a lot calculations.
Regarding the two movements in x and in y, these are:
-x (t): it is a uniform rectilinear motion.
-y (t): corresponds to a uniformly accelerated rectilinear motion with g = 9.8 m / s 2 and pointing vertically downwards.
In mathematical form:
The position vector is:
r (t) = i + j
In these equations the attentive reader will notice that the minus sign is due to gravity pointing towards the ground, the direction chosen as negative, while upwards is taken as positive.
Since velocity is the first derivative of position, simply differentiate r (t) with respect to time and obtain:
v (t) = v o cos α i + (v o. sin α - gt) j
Finally, the acceleration is expressed vectorially as:
a (t) = -g j
- Trajectory, maximum height, maximum time and horizontal reach
Trajectory
To find the explicit equation of the trajectory, which is the curve y (x), we must eliminate the time parameter, solving in the equation for x (t) and substituting in y (t). The simplification is somewhat laborious, but finally you get:
Maximum height
The maximum height occurs when v y = 0. Knowing that there is the following relationship between position and the square of the velocity:
Figure 3. The speed in the parabolic shot. Source: Giambattista, A. Physics.
Making v y = 0 just when reaching the maximum height:
With:
Maximum time
The maximum time is the time it takes the object to reach and max. To calculate it is used:
Knowing that v y becomes 0 when t = t max, it results:
Maximum horizontal reach and flight time
The range is very important, because it signals where the object will fall. This way we will know whether or not it hits the target. To find it we need the flight time, total time or v.
From the above illustration it is easy to conclude that t v = 2.t max. But beware! This is only true if the launch is level, that is, the height of the starting point is the same as the height of the arrival. Otherwise time is found by solving the quadratic equation that results from substituting the final and final position:
In any case, the maximum horizontal reach is:
Examples of parabolic shooting
The parabolic shot is part of the movement of people and animals. Also of almost all sports and games where gravity intervenes. For example:
Parabolic shooting in human activities
-The stone thrown by a catapult.
-The goal kick of the goalkeeper.
-The ball thrown by the pitcher.
-The arrow that comes out of the bow.
-All kinds of jumps
-Throw a stone with a sling.
-Any throwing weapon.
Figure 4. The stone thrown by the catapult and the ball kicked in goal kick are examples of parabolic shots. Source: Wikimedia Commons.
The parabolic shot in nature
-The water that flows from natural or artificial jets such as those from a fountain.
-Stones and lava gushing out of a volcano.
-A ball that bounces off the pavement or a stone that bounces on water.
-All kinds of animals that jump: kangaroos, dolphins, gazelles, cats, frogs, rabbits or insects, to name a few.
Figure 5. The impala is capable of jumping up to 3 m. Source: Wikimedia Commons. Arturo de Frias Marques / CC BY-SA (https://creativecommons.org/licenses/by-sa/3.0).
Exercise
A grasshopper jumps at an angle of 55º with the horizontal and lands 0.80 meters ahead. Find:
a) The maximum height reached.
b) If he jumped with the same initial speed, but forming an angle of 45º, would he go higher?
c) What can be said about the maximum horizontal reach for this angle?
Solution to
When the data supplied by the problem do not contain the initial velocity v or the calculations are somewhat more laborious, but from the known equations, a new expression can be derived. Starting from:
When it lands later, the height returns to 0, so:
Since t v is a common factor, it simplifies:
We can solve for t v from the first equation:
And replace in the second:
When multiplying all the terms by v or.cos α the expression is not altered and the denominator disappears:
Now you can clear v or o also substitute the following identity:
sin 2α = 2 sin α. cos α → v or 2 sin 2α = gx max
Calculate v or 2:
The lobster manages to maintain the same horizontal speed, but by decreasing the angle:
Reaches a lower height.
Solution c
The maximum horizontal reach is:
Changing the angle also changes the horizontal reach:
x max = 8.34 sin 90 / 9.8 m = 0.851 m = 85.1 cm
The jump is longer now. The reader can verify that it is maximum for the 45º angle because:
sin 2α = sin 90 = 1.
References
- Figueroa, D. 2005. Series: Physics for Sciences and Engineering. Volume 1. Kinematics. Edited by Douglas Figueroa (USB).
- Giambattista, A. 2010. Physics. Second Edition. McGraw Hill.
- Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed Prentice Hall.
- Resnick, R. 1999. Physics. Vol. 1. 3rd Ed. In Spanish. Compañía Editorial Continental SA de CV
- Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1.