In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. This is the principle of
compound motion established by
Galileo in 1638, and used by him to prove the parabolic form of projectile motion. A ballistic trajectory is a parabola with homogeneous acceleration, such as in a space ship with constant acceleration in absence of other forces. On Earth the acceleration changes magnitude with altitude as g(y)=g_0/(1+y/R)^2 and direction (faraway targets) with latitude/longitude along the trajectory. This causes an
elliptic trajectory, which is very close to a parabola on a small scale. However, if an object was thrown and the Earth was suddenly replaced with a
black hole of equal mass, it would become obvious that the ballistic trajectory is part of an elliptic
orbit around that "black hole", and not a parabola that extends to infinity. At higher speeds the trajectory can also be circular (
cosmonautics at
LEO?,
geostationary satellites at 5\frac{5}{6} R), parabolic or
hyperbolic (unless distorted by other objects like the Moon or the Sun). In this article a
homogeneous gravitational acceleration (g=g_0) is assumed.
Acceleration Since there is acceleration only in the vertical direction, the velocity in the horizontal direction is constant, being equal to \mathbf{v}_0 \cos\theta . The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to g.{{NoteTag|g is the
acceleration due to gravity. (9.81\,\mathrm{m/s^2} near the surface of the Earth).}} The components of the acceleration are: : a_x = 0 , : a_y = -g .*
*The y acceleration can also be referred to as the force of the earth (-F_g/m)
on the object(s) of interest. Velocity Let the projectile be launched with an initial
velocity \mathbf{v}(0) \equiv \mathbf{v}_0 , which can be expressed as the sum of horizontal and vertical components as follows: : \mathbf{v}_0 = v_{0x}\mathbf{\hat x} + v_{0y}\mathbf{\hat y} . The components v_{0x} and v_{0y} can be found if the initial launch angle θ is known: : v_{0x} = v_0\cos(\theta), : v_{0y} = v_0\sin(\theta) The horizontal component of the
velocity of the object remains unchanged throughout the motion. The vertical component of the velocity changes linearly, because the acceleration due to gravity is constant. The accelerations in the x and y directions can be integrated to solve for the components of velocity at any time t, as follows: : v_x = v_0 \cos(\theta) , : v_y = v_0 \sin(\theta) - gt . The magnitude of the velocity (under the
Pythagorean theorem, also known as the triangle law): : v = \sqrt{v_x^2 + v_y^2 } .
Displacement At any time t , the projectile's horizontal and vertical
displacement are: : x = v_0 t \cos(\theta) , : y = v_0 t \sin(\theta) - \frac{1}{2}gt^2 . The magnitude of the displacement is: : \Delta r=\sqrt{x^2 + y^2 } . Consider the equations, : x = v_0 t \cos(\theta) and y = v_0 t\sin(\theta) - \frac{1}{2}gt^2. If t is eliminated between these two equations the following equation is obtained: : y = \tan(\theta) \cdot x-\frac{g}{2v^2_{0}\cos^2 \theta} \cdot x^2=\tan\theta \cdot x \left(1-\frac{x}{R}\right). Here R is the
range of a projectile. Since g, θ, and v0 are constants, the above equation is of the form : y=ax+bx^2 , in which a and b are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical. If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v0 in the afore-mentioned parabolic equation: : v_0 = \sqrt{{x^2 g} \over {x \sin 2\theta - 2y \cos^2\theta}} .
Displacement in polar coordinates The parabolic trajectory of a projectile can also be expressed in polar coordinates instead of
Cartesian coordinates. In this case, the position has the general formula : r( \phi ) = \frac{2v_0^2 \cos^2\theta} \left(\tan\theta\sec\phi -\tan\phi\sec\phi \right) . In this equation, the origin is the midpoint of the horizontal range of the projectile, and if the ground is flat, the parabolic arc is plotted in the range 0 \leq \phi \leq \pi . This expression can be obtained by transforming the Cartesian equation as stated above by y = r \sin\phi and x = r \cos\phi .
Time of flight or total time of the whole journey The total time t for which the projectile remains in the air is called the time-of-flight. : y=v_0 t\sin(\theta) -\frac{1}{2} g t^2 After the flight, the projectile returns to the horizontal axis (x-axis), so y=0 . : v_0 t\sin(\theta) =\frac{1}{2} g t^2 : v_0 \sin(\theta) = \frac{1}{2}gt : t=\frac{2 v_0\sin(\theta)}{g} Note that we have neglected air resistance on the projectile. If the starting point is at height y0 with respect to the point of impact, the time of flight is: : t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g} As above, this expression can be reduced (y0 is 0) to : t = \frac{v\sin{\theta} + \sqrt{(v\sin{\theta})^2}}{g} = \frac{2v\sin{\theta}}{g}, = \frac{2v\sin{(45^\circ)}} = \frac{\sqrt{2}v} if θ equals 45°.
Time of flight to the target's position As shown above in the
Displacement section, the horizontal and vertical velocity of a projectile are independent of each other. Because of this, we can find the time to reach a target using the displacement formula for the horizontal velocity: x = v_0 t \cos(\theta) \frac{x}{t}=v_0\cos(\theta) t=\frac{x}{v_0\cos(\theta)} This equation will give the total time
t the projectile must travel for to reach the target's horizontal displacement, neglecting air resistance.
Maximum height of projectile The greatest height that the object will reach is known as the peak of the object's motion. The increase in height will last until v_y=0 , that is, : 0=v_0 \sin(\theta) - gt_h . Time to reach the maximum height(h): : t_h = \frac{v_0 \sin(\theta)} . This matches the formula found above (the time of flight) as the maximum height (in a parabola) sits in the middle of its roots. This means the projectile reaches
h in the middle of the journey, therefore, in half of the time. For the vertical displacement of the maximum height of the projectile: : h = v_0 t_h \sin(\theta) - \frac{1}{2} gt^2_h : h = \frac{v_0^2 \sin^2(\theta)}{2|g|} The maximum reachable height is obtained for
θ=90°: : h_{\mathrm{max}} = \frac{v_0^2}{2|g|} If the projectile's position (x,y) and launch angle (θ) are known, the maximum height can be found by solving for h in the following equation: :h=\frac{(x\tan\theta)^2}{4(x\tan\theta-y)}. Angle of
elevation (φ) at the maximum height is given by: :\phi = \arctan
Relation between horizontal range and maximum height The relation between the range d on the horizontal plane and the maximum height h reached at \frac{t_d}{2} is: : h = \frac{d\tan\theta}{4} h = \frac{v_0^2\sin^2\theta}{2|g|} : d = \frac{v_0^2\sin2\theta} : \frac{h}{d} = \frac{v_0^2\sin^2\theta}{2|g|} × \frac{g}{v_0^2\sin2\theta} : \frac{h}{d} = \frac{\sin^2\theta}{4\sin\theta\cos\theta} h = \frac{d\tan\theta}{4} . If h = R :
\theta = \arctan(4)\approx 76.0^\circ Maximum distance of projectile The range and the maximum height of the projectile do not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction. The horizontal range d of the projectile is the horizontal distance it has traveled when it returns to its initial height (y=0). : 0 = v_0 t_d \sin(\theta) - \frac{1}{2}gt_d^2 . Time to reach ground: : t_d = \frac{2v_0 \sin(\theta)} . From the horizontal displacement the maximum distance of the projectile: : d = v_0 t_d \cos(\theta) , so : d = \frac{v_0^2}\sin(2\theta). Note that d has its maximum value when : \sin(2\theta)=1, which necessarily corresponds to 2\theta=90^\circ , or \theta=45^\circ . The total horizontal distance (d) traveled. : d = \frac{v \cos \theta} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right) When the surface is flat (initial height of the object is zero), the distance traveled: : d = \frac{v^2 \sin(2 \theta)} Thus the maximum distance is obtained if θ is 45 degrees. This distance is: : d_{\mathrm{max}} = \frac{v^2}
Application of the work energy theorem According to the
work-energy theorem the vertical component of velocity is: : v_y^2 = (v_0 \sin \theta)^2-2gy . These formulae ignore
aerodynamic drag and also assume that the landing area is at uniform height 0.
Angle of reach The "angle of reach" is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v. : \sin(2\theta) = \frac{gd}{v^2} There are two solutions: : \theta = \frac{1}{2} \arcsin \left( \frac{gd}{v^2} \right) (shallow trajectory) and because \sin(2\theta) = \cos (2\theta - 90^\circ ), : \theta = 45^\circ + \frac{1}{2} \arccos \left( \frac{gd}{v^2} \right) (steep trajectory)
Angle θ required to hit coordinate (x, y) To hit a target at range x and altitude y when fired from (0,0) and with initial speed v, the required angle(s) of launch θ are: : \theta = \arctan{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} The two roots of the equation correspond to the two possible launch angles, so long as they aren't imaginary, in which case the initial speed is not great enough to reach the point (x,y) selected. This formula allows one to find the angle of launch needed without the restriction of y= 0 . One can also ask what launch angle allows the lowest possible launch velocity. This occurs when the two solutions above are equal, implying that the quantity under the
square root sign is zero. This, tan θ = v2/gx, requires solving a
quadratic equation for v^2 , and we find : v^2/g = (2gy\pm g\sqrt{4y^2+4(1x)^2})/2g = y\pm\sqrt{y^2+x^2}. This gives : \theta=\arctan\left(y/x+\sqrt{y^2/x^2+1}\right). If we denote the angle whose tangent is by , then : \tan\theta=\frac{\sin\alpha+1}{\cos\alpha}, its reciprocal: : \tan(\pi/2-\theta)=\frac{\cos\alpha}{\sin\alpha+1}, : \cos^2(\pi/2-\theta)=\frac 12(\sin\alpha+1) : 2\cos^2(\pi/2-\theta)-1=\cos(\pi/2-\alpha) This implies : \theta = \pi/2 - \frac 12(\pi/2-\alpha). In other words, the launch should be at the angle (\pi/2 + \alpha)/2 halfway between the target and
zenith (vector opposite to gravity).
Total Path Length of the Trajectory The length of the parabolic arc traced by a projectile, L, given that the height of launch and landing is the same (there is no air resistance), is given by the formula: :L = \frac{v_0^2}{2g} \left( 2\sin\theta + \cos^2\theta\cdot\ln \frac{1 + \sin\theta}{1 - \sin\theta} \right) = \frac{v_0^2}{g} \left( \sin\theta + \cos^2\theta\cdot\tanh^{-1}(\sin\theta) \right) where v_0 is the initial velocity, \theta is the launch angle and g is the acceleration due to gravity as a positive value. The expression can be obtained by evaluating the
arc length integral for the height-distance parabola between the bounds
initial and
final displacement (i.e. between 0 and the horizontal range of the projectile) such that: :L = \int_{0}^{\mathrm{range}} \sqrt{1 + \left ( \frac{\mathrm{d}y}{\mathrm{d}x} \right )^2}\,\mathrm{d}x = \int_{0}^{v_0^2 \sin(2\theta)/g} \sqrt{1+\left (\tan\theta -{g\over {v_0^2 \cos^2\theta}}x\right)^2}\,\mathrm{d}x . If the time-of-flight is
t, :L = \int_{0}^{t} \sqrt{v_x^2 + v_y^2}\,\mathrm{d}t = \int_{0}^{2v_0\sin\theta/g} \sqrt{(gt)^2-2gv_0\sin\theta t+v_0^2}\,\mathrm{d}t. == Trajectory in air ==