When they study mechanical motion in physics, then after familiarizing themselves with the uniform and uniformly accelerated movement of objects, they proceed to consider the motion of the body at an angle to the horizontal. In this article we will study this issue in more detail.
What is the movement of the body at an angle to the horizon?
This type of movement of objects occurs when a person throws a stone into the air, a gun fires a core, or a goalkeeper knocks a soccer ball out of the goal. All such cases are examined by the science of ballistics.
The marked type of movement of objects in the air occurs along a parabolic trajectory. In the general case, carrying out the corresponding calculations is not easy, since it is necessary to take into account the air resistance, the rotation of the body during the flight, the rotation of the Earth around the axis and some other factors.
In this article, we will not take into account all these factors, but consider the issue from a purely theoretical point of view. Nevertheless, the obtained formulas describe quite well the trajectories of bodies moving over small distances.
Obtaining formulas for the considered type of motion
We derive the formulas for the motion of the body to the horizon at an angle. In this case, we will take into account only one single force acting on a flying object - gravity. Since it acts vertically downward (parallel to and against the y axis), then, considering the horizontal and vertical components of the movement, we can say that the former will have the character of uniform rectilinear movement. And the second - equally slow (uniformly accelerated) rectilinear movement with acceleration g. That is, the components of the velocity through the value of v 0 (initial velocity) and θ (angle of the direction of movement of the body) are written as follows:
v x = v 0 * cos (θ)
v y = v 0 * sin (θ) -g * t
The first formula (for v x ) is always valid. As for the second one, one nuance must be noted here: the minus sign in front of the product g * t is placed only if the vertical component v 0 * sin (θ) is directed upwards. In most cases, this happens, however, if you throw the body from a height, pointing it down, then in the expression for v y you should put the "+" sign in front of g * t.
Having integrated the formulas for the velocity components over time, and taking into account the initial height h of the body’s flight, we obtain the equations for the coordinates:
x = v 0 * cos (θ) * t
y = h + v 0 * sin (θ) * tg * t 2/2
Range calculation
When considering in physics the motion of a body toward the horizon at an angle, useful for practical application, it turns out the calculation of the flight range. Define it.
Since this movement is a uniform movement without acceleration, it is enough to substitute the flight time into it and get the desired result. Flight range is determined solely by movement along the x axis (parallel to the horizon).
The time spent by the body in the air can be calculated by equating the y coordinate to zero. We have:
0 = h + v 0 * sin (θ) * tg * t 2/2
This quadratic equation is solved through the discriminant, we obtain:
D = b 2 - 4 * a * c = v 0 2 * sin 2 (θ) - 4 * (- g / 2) * h = v 0 2 * sin 2 (θ) + 2 * g * h,
t = (-b ± √ D) / (2 * a) = (-v 0 * sin (θ) ± √ (v 0 2 * sin 2 (θ) + 2 * g * h)) / (- 2 * g / 2) =
= (v 0 * sin (θ) + √ (v 0 2 * sin 2 (θ) + 2 * g * h)) / g.
In the last expression, one root with a minus sign is discarded, in view of its insignificant physical meaning. Substituting the flight time t in the expression for x, we obtain the flight range l:
l = x = v 0 * cos (θ) * (v 0 * sin (θ) + √ (v 0 2 * sin 2 (θ) + 2 * g * h)) / g.
The easiest way to analyze this expression is if the initial height is zero (h = 0), then we get a simple formula:
l = v 0 2 * sin (2 * θ) / g
This expression indicates that the maximum flight range can be obtained if the body is thrown at an angle of 45 o (sin (2 * 45 o ) = m1).
Maximum body lift
In addition to the flight range, it is also useful to find the altitude above which the body can climb. Since this type of movement is described by a parabola, the branches of which are directed downward, the maximum height of rise is its extremum. The latter is calculated by solving the equation for the derivative with respect to t for y:
dy / dt = d (h + v 0 * sin (θ) * tg * t 2/2) / dt = v 0 * sin (θ) -gt = 0 =>
=> t = v 0 * sin (θ) / g.
Substituting this time in the equation for y, we get:
y = h + v 0 * sin (θ) * v 0 * sin (θ) / gg * (v 0 * sin (θ) / g) 2/2 = h + v 0 2 * sin 2 (θ) / ( 2 * g).
This expression indicates that the body will rise to the maximum height if it is thrown vertically upward (sin 2 (90 o ) = 1).