The movement of the body thrown at an angle to the horizon is described in two coordinates. One characterizes flying range, another – height. Time of flight depends on the maximum height which is reached by a body.
Instruction
1. Let the body be thrown at an angle α to the horizon with an initial speed of v0. Initial coordinates of a body let will be zero: x (0)=0, y(0)=0. In projections to coordinate axes the initial speed will decay on two components: v0(x) and v0(y). Same belongs to function of speed in general. On Ox axis the speed conditionally is considered a constant, on an axis of Oy changes under the influence of gravity. Acceleration of gravity of g can be accepted approximately for 10m/with².
2. A corner α under which the body is thrown, it is set not accidentally. Through it it is possible to paint initial speed in coordinate axes. So, v0(x) =v0·cos(α), v0(y) =v0·sin(α). Now it is possible to receive function of coordinate components of speed: v(x) =const=v0 (x) =v0·cos(α), v(y) =v0 (y)-g · t=v0 · sin(α)-g·t.
3. Coordinates of a body x and y depend on t time. Thus, it is possible to work out two equations of dependence: x=x0+v0(x) · t+a(x) · t²/2, y=y0+v0(y) · t+a(y) · t²/2. As on a condition of x0=0, a(x)=0, x=v0(x) · t=v0 · cos(α)·t. It is also known that y0=0, a (y)=-g (the sign "minus" appears because that the direction of acceleration of gravity of g and the positive direction of an axis Oy are opposite). Therefore y=v0 · sin(α) · t-g · t²/2.
4. Time of flight can be expressed from a speed formula, knowing that in the maximum point the body for a moment stops (v=0), and are equal to duration of "rise" and "descent". So, at substitution of v(y) =0 in the equation of v(y) =v0·sin(α)-g·t it turns out: 0=v0·sin(α)-g·t(p) where t(p) is peak time, "topmost t". From here t(p) =v0·sin(α)/g. The general time of flight then will be expressed as t=2 · v0 · sin(α)/g.
5. The same formula can be received and in a different way, mathematical, from the equation for y=v0 coordinate · sin(α) · t-g · t²/2. This equation can be rewritten in a little changed look: y=-g/2 · t²+v0·sin(α)·t. It is visible that it is square dependence where y is function, t is an argument. Top of the parabola describing a trajectory is the point of t(p)= [-v0 · sin(α)] / [-2g/2]. Minuses and the two are reduced therefore t(p) =v0·sin(α)/g. If to designate the maximum height for H and to remember that the peak point is top of a parabola on which the body, then H=y (t(p)) =v0²sin² moves (α) / 2g. That is, to receive height, it is necessary to substitute "topmost t" in the equation for y coordinate.
6. So, time of flight registers as t=2 · v0 · sin(α)/g. That to change it, it is necessary to change the initial speed and a tilt angle respectively. The more speed – the longer flies a body. With a corner it is slightly more difficult, time depends not on the corner, and on its sine. The greatest possible value of a sine – unit – is reached at a tilt angle in 90 °. It means that most longer the body flies when it is thrown vertically up.
7. Flying range is final coordinate x. If to substitute the flight time found already in x=v0 equation · cos(α) · t, it is easy to find that L=2v0²sin(α) cos(α)/g. Here it is possible to apply a trigonometrical formula of a double corner 2sin(α) cos(α)=sin(2α), then L=v0²sin(2α)/g. The sine two an alpha is equal to unit when 2α = p/2, α = p/4. Thus, flying range is maximum in case to throw a body at an angle 45 °.