Parabola_Water.jpgA particle is projected in a vertical plane with an initial velocity of at an angle of α with the horizontal. Any air resistance and variation in the force of gravity is neglected.

Thus only a constant force of gravity is acting on the particle vertically downwards.


R = Range of projectile
H = Maximum height of projectile

The best way to study this motion is to consider the horizontal and vertical motions separately.

Horizontally there is no acceleration since no horizontal force acts on the particle. 
The initial horizontal velocity is vcosα


But x = 0 when t = 0 therefore, c = 0

Therefore, x = vcosαt ...............A

Vertically there is a constant downwards acceleration of g.

Note: is approximately 10 m/s2

Therefore, Y12_Projectile_Motion_03.gif

Note: The minus is needed to represent the downward direction.

Therefore, Y12_Projectile_Motion_04.gif

Initially, the vertical velocity is Vsinα










The equations A and B enable us to find the position of the projectile at any time t.

The time t is a PARAMETER and (x, y) are PARAMETRIC EQUATIONS enabling the equation of the path traced out by the projectile in the x-y plane to be found.





From the basic equations we can deduce a variety of useful measures.

Greatest Height Reached







When y = 0, solve C for x



We further note that the maximum value that sin2α can have is 1 when α = 45o

Therefore, To obtain the MAXIMUM range we project at 45o

and Y12_Projectile_Motion_15.gif





An arrow is fired at an angle of 60o to the horizontal with an initial velocity of 50 m/s.
a. Find the range, maximum height and time of flight of the arrow.
b. With the same initial velocity, what is the maximum range of this arrow?
Take g = 10 m/s2

a. Here V = 50, α = 60 and g = 10


Time of flight T is given by solving y = 0

Therefore, 0 = 50sin60.T − 1/2. 10. T2

T = 10sin 60 = 5√3 = 8.7 seconds

b. For maximum range

RMAX = V2 / g = 2500/10

Therefore, RMAX = 250 metres