Projectile Equations with Explanations
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.
The 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}
The shape of the trajectory followed by the projectile is found as follows
Solve the formula x = V_{0} cos(θ) t for t to obtain
Substitute t by x / V_{0} cos(θ) in the expression of y obtained above
y =

V_{0} sin(θ) x
V_{0} cos(θ)
_{ (1/2)g}


Simplify
y =

 g
2 [ V_{0} cos(θ) ]^{2}
x^{2} + tan(θ) x

The above equation is that of a parabola of the form
y = A x^{2} + B x
where

A =

 g
2 [ V_{0} cos(θ) ]^{2}

and B = tan(θ)

Path of Projectile: y = A x^{2} + B x
A =

 g
2 [ V_{0} cos(θ) ]^{2}

and B = tan(θ)

Time of flight of the projectile
The 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
t(V_{0} sin(θ) t  (1/2) g) = 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
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. Hence
V_{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 = V_{0} sin(θ) V_{0} sin(θ) / g  (1/2) g [ V_{0} sin(θ) / g ]^{2}
Maximum Height (at point B) =



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. Hence
range = AC = x(2 V_{0} sin(θ) / g)
= V_{0} cos(θ) 2 V_{0} sin(θ) / g
= V_{0}^{2} sin(2θ) / g
Horizontal Range = V_{0}^{2} sin(2θ) / g

