# Projectile Equations with Explanations

Consider a projectile being launched at an initial velocity v0 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/s2. 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 V0 being a vector quantity, has two components:

V0x and V0y given by

V0x = V0 cos(θ)

V0y = V0 sin(θ)

The acceleration A is a also a vector with two components Ax and Ay given by

Ax = 0 and Ay = - g = - 9.8 m/s2

Along the x axis the acceleration is equal to 0 and therefore the velocity Vxis constant and is given by

Vx = V0 cos(θ)

Along the y axis, the acceleration is uniform and equal to -g and the velocity at time t is given by

Vy = V0 sin(θ) - g t

Along the x axis the velocity Vxis constant and therefore the component x of the displacement is given by

x = V0 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 = V0 sin(θ) t - (1/2) g t2

The shape of the trajectory followed by the projectile is found as follows

Solve the formula x = V0 cos(θ) t for t to obtain

 t = x V0 cos(θ)

Substitute t by x / V0 cos(θ) in the expression of y obtained above

 y = V0 sin(θ) x V0 cos(θ)   - (1/2)g x2 (V0 cos(θ))2

Simplify

 y = - g 2 [ V0 cos(θ) ]2  x2 + tan(θ) x

The above equation is that of a parabola of the form

y = A x
2 + B x

 where A = - g 2 [ V0 cos(θ) ]2 and B = tan(θ)
Path of Projectile: y = A x2 + B x
 A = - g 2 [ V0 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 = V0 sin(θ) t - (1/2) g t2 = 0

t(V0 sin(θ) t - (1/2) g) = 0

Two solutions: t = 0 (correspond to point A) and t = 2 V0 sin(θ) / g (correspond to point C)

Hence the time of flight = 2 V0 sin(θ) / g

Time of Flight = 2 V0 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 = V0 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) V0 sin(θ) / g - (1/2) g [ V0 sin(θ) / g ]2

 = [ V0 sin(θ) ]2 2 g

 Maximum Height (at point B) = [ V0 sin(θ) ]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 V0 sin(θ) / g obtained above. Hence

range = AC = x(2 V0 sin(θ) / g)

= V0 cos(θ) 2 V0 sin(θ) / g

= V02 sin(2θ) / g

Horizontal Range = V02 sin(2θ) / g