Consider a projectile being launched at an initial velocity v_{0} in a direction making an angle θ with the horizontal. We assume that air resistance is negligible and the only force acting on the object is the force of gravity with acceleration g = 9.8 m/s^{2}. Also an interactive html 5 applet may be used to better understand the projectile equations. Projectile problems with solutions are also included in this site.
Velocity, Acceleration and Displacement of the ProjectileThe initial velocity V_{0} being a vector quantity, has two components:V_{0x} and V_{0y} given by V_{0x} = V_{0} cos(θ) V_{0y} = V_{0} sin(θ) The acceleration A is a also a vector with two components A_{x} and A_{y} given by A_{x} = 0 and A_{y} =  g =  9.8 m/s^{2} Along the x axis the acceleration is equal to 0 and therefore the velocity V_{x} is constant and is given by V_{x} = V_{0} cos(θ) Along the y axis, the acceleration is uniform and equal to  g and the velocity at time t is given by V_{y} = V_{0} sin(θ)  g t Along the x axis the velocity V_{x}is constant and therefore the component x of the displacement is given by x = V_{0} cos(θ) t Along the y axis, the motion is that of a uniform acceleration type and the y component of the displacement is given by y = V_{0} sin(θ) t  (1/2) g t^{2}
Shape of the Trajectory of the ProjectileThe shape of the trajectory followed by the projectile is found as followsSolve the formula \( \; x = V_0 cos(\theta) t \; \) for \( t \) to obtain \[ t = \dfrac{x}{V_0 cos(\theta)} \] Substitute \( t \) by \( \dfrac{x}{V_0 cos(\theta)} \) in the above expression of y to obtain \[ y = \dfrac{V_0 \sin(\theta) x}{V_0 \cos(\theta)}  (1/2) g \left( \dfrac{x}{V_0 \cos(\theta)} \right)^2 \] Simplify \[ y =  \dfrac{g \; x^2}{2( V_0 \cos(\theta))^2} + x \tan(\theta) \] The above equation is the path of projectile which is a parabola of the form \( y = A x^2 + B x \) where \( A =  \dfrac{g}{2( V_0 \cos(\theta))^2} \) and \( B = \tan(\theta) \)
Time of Flight of the ProjectileThe time of flight is the time taken for the projectile to go from point A to point C (see figure above).It is calculated by setting y = 0 (y = 0 at point C) and solve for t y = V_{0} sin(θ) t  (1/2) g t^{2} = 0 Factor t out in the above equation t(V_{0} sin(θ)  (1/2) g t) = 0 Two solutions: t = 0 (correspond to point A) and t = 2 V_{0} sin(θ) / g (correspond to point C) Hence the time of flight = 2 V_{0} sin(θ) / g
Maximum Height of the Projectile (corresponding to point B in figure above)At point B in the figure above, the projectile is momentarily horizontal and therefore the vertical component of its velocity is equal to zero. HenceV_{y} = V_{0} sin(θ)  g t = 0 Solve for t to obtain t = V_{0} sin(θ) / g (Note: this is half the time of flight because of the symmetry of the parabola) Substitute t by V_{0} sin(θ) / g in the expression of y, we obtain the maximum height \( H = \dfrac{V_0 \sin(\theta) V_0 sin(\theta)}{g}  (1/2) \dfrac{}{} g \left( \dfrac{V_0 sin(\theta)}{g} \right)^2 = \dfrac{(V_0 sin(\theta))^2}{2 g} \) Horizontal Range of a Projectile (distance AC in the figure above)Distance AC which is the horizontal range is equal to x when t is equal to the time of flight 2 V_{0} sin(θ) / g obtained above. Hencerange AC = x = V_{0} cos(θ) t at t = time of flight = 2 V_{0} sin(θ) / g Substitute t by 2 V_{0} sin(θ) / g and simplify to obtain the range AC AC \( = V_0 cos(\theta) 2 V_0 sin(\theta) / g = V_0^2 sin(2\theta) / g \) More References and linksInteractive Simulation of Projectile.Projectile problems with solutions.
