When considering the movement of bodies in free space, whether it is the movement of planets in space or objects in the Earth’s atmosphere, gravitational forces are the defining trajectory. This article discusses one of the types of such movement - movement at an angle to the horizon. The formulas necessary to solve the corresponding problems are also given.
What type of movement will be discussed?
Movement at an angle to the horizon (formulas are given later in the article) is the movement of objects in the gravitational field of our planet, during which only one single force acts - gravity. In reality, air resistance is still present, but it is customary to neglect it.
In the process of moving at an angle to the horizon, the body begins to move either from the surface of the earth, or from a certain height h. At the initial time, it has a velocity v directed at a certain angle to the horizon. Flying along a certain trajectory, the body falls to the ground.
The marked trajectory is considered to be parabolic, but this is not entirely true. The fact is that, as a result of the action of gravity, the object moves along an elliptical trajectory (due to the fall to the ground, the body describes only part of the ellipse). In some cases, part of the real elliptical trajectory can be represented as a parabola with high accuracy, which is done in order to simplify the mathematical calculations in the calculations.
Features of movement at an angle to the horizon
Since in the process of such movement of objects only one gravity acts, which is constant over the entire path (always directed down and has a constant absolute value), this type of movement has the following properties:
- Knowing the initial speed, the angle to the horizon and the value of the height from which the body starts, you can unambiguously calculate its entire flight path.
- An increase in the initial velocity at a constant angle to the horizon leads to an increase in the flight range.
- If the body starts flying from zero height, then the angle of its fall will be exactly equal to the angle of departure.
- The horizontal and vertical movements of the body during movement are independent, so they can be analyzed separately from each other.
Examples of movement type in question
The flight of a ball when it is hit by a soccer player or the movement of a projectile in the air after they have been shot from any weapon (gun, mortar, tank) are the most striking examples of movement at an angle to the horizon. Also such is the flight of a stone thrown by a hand from any height, or the jumping of the same stone when it is repelled from water.
In contrast to the above examples, take-off, flight and landing of the aircraft do not apply to the movement in question, since in this case additional forces act in addition to gravity (thrust, wing lift).
Physics of motion at an angle to the horizon: formulas
As mentioned above, when the body begins its flight, then two independent equations of its motion along the horizon and perpendicular to it can be compiled. First, we write down what forces act on the body in the x and y directions:
F x = 0;
F y = -m * g.
Here m is the mass of the body, g is the acceleration that our planet informs all the bodies near its surface. A minus sign indicates that gravity F y acts against the direction of the y axis.
Now we write the components of the initial velocity on each axis:
v x = v * cos (θ);
v y = v * sin (θ).
Here θ is the angle to the horizon. Since the acting force leads to a change in speed according to Newton’s second law, we can write:
v x = const.
v y = v * sin (θ) - g * t.
Here t is the point in time after the start of flight.
Integrating both expressions over time, we obtain the final formulas for motion at an angle to the horizon:
x = v x * t + x 0 = v * cos (θ) * t + x 0 ;
y = v y * t + y 0 = v * sin (θ) * t - g * t 2/2 + y 0 .
Two constants appeared in the resulting expressions: x 0 and y 0 . They describe the initial coordinates of the object. When mathematically solve the problem of movement along a parabolic trajectory, then x 0 is assumed to be equal to zero (reference). As for y 0 , the situation here is a little more complicated: if the body starts from the surface of the earth, then y 0 is also zero; if the body starts moving from a height h, then y 0 is equal to this height. Thus, the motion formulas at an angle to the horizon from a height will take the form:
x = v * cos (θ) * t.
y = v * sin (θ) * t - g * t 2/2 + h.
Problem solving example
We will solve an interesting problem of body motion at an angle to the horizon. Its condition is as follows: an archer, being on a tower 15 meters high, launches an arrow at a speed of 20 m / s at an angle of 0 o to the surface of the earth. You must determine how far from the tower the arrow will fly.
The desired distance will be equal to the change in the x coordinate, that is:
x = v * cos (θ) * t = v * t.
The cosine of the zero angle is unity, so it was omitted. So, to get an answer, you need to find the boom flight time t. To do this, we turn to the second equation (along the y axis). According to the condition of the problem, it has the form:
y = v * sin (θ) * t - g * t 2/2 + h = h - g * t 2/2, since sin (0) = 0.
When the arrow falls to the ground, its y coordinate will become 0, so we get the equation:
h - g * t 2/2 = 0.
This equality is called a pure second-order equation. It is solved by transferring the free term to another part of the equality, using the square root, that is:
t = √ (2 * h / g).
It remains to substitute this formula in the equation for x and get the desired answer:
x = v * √ (2 * h / g) = 20 * √ (2 * 15 / 9.81) = 34.97 ≈ 35 meters.
Thus, the arrow will fly away only 35 meters. The archer can increase the range of her flight if she directs the arrow at an angle θ to a non-zero horizon.