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 \( \; x = V_0 cos(\theta) t \; \) 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 \( B = \tan(\theta) \)

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

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

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

\( H = \dfrac{V_0 \sin(\theta) V_0 sin(\theta)}{g} - (1/2) \dfrac{}{} g \left( \dfrac{V_0 sin(\theta)}{g} \right)^2 = \dfrac{(V_0 sin(\theta))^2}{2 g} \)

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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 = 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 \( = V_0 cos(\theta) 2 V_0 sin(\theta) / g = V_0^2 sin(2\theta) / g \)