Parabolic motion:

Parabolic motion

Projectiles in flight, sums of rectilinear motions, and range have in common that they are explained by the Parabolic Motion and its formulas. In this new article, we see everything there is to know about the formulaslaws, and examples of this new member of the kinematic physics family. I’d start with a definition. what do you think?

Parabolic motion and its formulas describe the movement of an object that follows a parabolic trajectory under the effect of gravity .

This type of motion occurs when an object is thrown with an initial horizontal velocity and a vertical component.
It can be considered as the combination of two separate motions along the horizontal and vertical axes. The object moves with a constant speed along the horizontal axis and undergoes uniformly accelerated motion along the vertical axis due to gravity.

Common examples of parabolic motion are the motion of a projectile , or the motion of a soccer ball. We will see them in the next scrolls, so let’s dive into this new chapter dedicated to kinematic physics .

What is parabolic motion and what do the formulas mean

Let’s start by saying that parabolic motion is a type of displacement that occurs when an object is thrown with an initial horizontal velocity and a vertical component. Both are independent movements , which combine to produce the overall parabolic trajectory of motion.

During the creation of the parabola the object undergoes two types of acceleration:

  • the downward gravitational acceleration ;
  • the horizontal acceleration , which is zero .

Both influence the speed and position of the object over time and produce the overall parabolic trajectory of the motion.The horizontal component of the velocity remains constant throughout the movement, while the vertical component of the velocity varies due to gravitational acceleration.

In fact, an object:

  • it reaches its maximum height when its vertical speed vanishes, therefore starting to descend due to gravitational acceleration;
  • it descends along the parabolic trajectory until it touches the ground (or the surface from which it started).
  • Along the horizontal axis we have constant velocity due to uniformly accelerated motion, while on the vertical axis we find the force of gravity.

This allows us to understand that parabolic motion calculates various parameters, such as the maximum height reached, the horizontal distance traveled and the speed of the object at different instants of time.
Each of these is expressed by the main formulas used in parabolic motion.

Parabolic formulas and projectiles

Ok, but how is this type of movement calculated? The answer lies in the formulas of parabolic motion. The main formulas used in parabolic motion are the following.

The first we will see is the hourly law of parabolic motion , useful for calculating the height (y) as a function of time (t):
y = v₀ v t – (1/2)gt²
where:

  • y represents the vertical position of the object (height) at time t;
  • v₀ v is the vertical component of the initial velocity;
  • g is the gravitational acceleration (usually taken as 9.8 m/s²);
  • t is the elapsed time.

This formula takes into account the downward gravitational acceleration, which causes the object’s vertical position to change over time.

The first term of the formula represents the vertical distance traveled by the object in time t, while the second term represents the change in height caused by gravitational acceleration. Try to notice that the hourly law of parabolic motion describes only the vertical component of the position.

The hourly law formula of parabolic motion describes only vertical motion.  A rocket, like the one in the photo, follows the same parabolic trajectory as a projectile.

To determine the horizontal position of the object, it is necessary to consider the motion along the horizontal axis .

Formula for horizontal distance (x) traveled:
x = v₀ₓt
where v₀ₓ represents the horizontal component of the initial velocity and t is the elapsed time.

This formula gives the horizontal distance traveled by the object in time t.
Since there is no horizontal acceleration in parabolic motion, the horizontal velocity remains constant and equal to the horizontal component of the initial velocity. The combination of the hourly laws for motion along the horizontal and vertical axis allows the parabolic trajectory of the moving object to be completely described.

Parabolic motion and formulas for the trajectory parameters

Furthermore, there are two other formulas to consider to calculate various parameters of parabolic motion, such as the maximum height reached, the horizontal distance traveled, and the speed of the object at different instants of time.

Formula for horizontal velocity (vₓ) :
vₓ = v₀ₓ
where v₀ₓ represents the horizontal component of the initial velocity.

This indicates that the horizontal velocity remains constant throughout the motion, since there are no forces acting horizontally on the object.

Formula for vertical velocity (vᵥ) as a function of time (t) :
vᵥ = v₀ᵥ – gt
where v₀ᵥ represents the vertical component of the initial velocity, g is the gravitational acceleration and t is the elapsed time.

This formula describes how the object’s vertical velocity varies over time due to gravitational acceleration. As mentioned, vertical velocity decreases linearly over time due to downward acceleration.

Projectile motion: examples and exercises

A classic example of parabolic motion and the application of its formulas is the motion of a projectile (or a balloon). In short, the motion of the projectile can be described as a parabolic trajectory. In fact, the object follows a trajectory that is symmetrical with respect to the horizontal axis and open upwards, reaching a maximum height before returning to the ground.

This motion is governed by parabolic motion formulas, which provide us with a mathematical understanding of how the projectile moves in its flight. Let’s see an exercise on projectile motion that uses the parabolic motion formulas.

A ball is thrown vertically upwards at a speed of 19.6 m/s. What distance did he travel in 2 seconds? Choose one of the following:

  • 0m;
  • 9.8m;
  • 14.7m;
  • 19.6m;
  • 39.2 m

Don’t worry, we’ll solve it together. To solve the exercise, we can use the formula y = v₀ v t – (1/2)gt² .

Where y represents the vertical position (height) of the object, v₀ᵥ is the vertical component of the initial velocity, t is the elapsed time, and g is the gravitational acceleration (generally taken to be 9.8 m/s²).

In our case, the initial vertical velocity v₀ᵥ is 19.6 m/s and the time t is 2 seconds. Substituting these values ​​into the formula, we get:

y = (19.6 m/s) * (2 s) – (1/2) * (9.8 m/s²) * (2 s)²

Calculating the result, we get:

y = 39.2 m – 19.6 m = 19.6 m

So, the distance the ball travels in 2 seconds is 19.6 meters.

The correct answer is therefore: 19.6 m.

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