Projectiles
When a body is projected from
a point in such a way that the only force assumed to be acting is gravity then
that body is classed as a projectile.
Parametric and Cartesian forms of equations of the trajectory (flight path)
The derivation of the following equations
is a little tricky but the process has appeared on an Edexcel
paper at M2.
Suppose a particle P is projected from O at angle α and is at the point (x,y) at time t after leaving O.
Equation of motion for P horizontally is F
= ma
Therefore 
and velocity parallel
to Ox is constant and equal to Vcosα.
For constant velocity x
= vel × time
x
= Vcosαt (1)
Equation of motion for P
vertically F = ma
Using constant acceleration equations 
With u = sinα and a = -g 
(2)
1 and 2 are the parametric equations of trajectory.
Rearrange 1 to give:
Substitute into equation 2
(3)
Equation 3 is the Cartesian form of the trajectory of a
projectile. Assuming V, g and α are
constants for any particular case, this is the equation of a parabola.
Maximum height of a projectile.
At the maximum height the vertical
component of the velocity is zero.
Time to maximum height and time of flight
Using v
= u + at
At maximum height v = 0
u = Vsinα 0 = Vsinα – gt
For time of flight on the horizontal plane we use:
There is zero displacement vertically hence s = 0.
and 
By symmetry this time is twice the value to maximum height.
For sloping planes, (cliffs etc) use s as the vertical
displacement from the starting point to the landing point.
Range on the horizontal plane
Using equation 1 x
= Vcosαt
Time in flight is
Therefore range R is: - 
2sinαcosα =
sin2α
Maximum range appears when sin2α = 1
Therefore
α = 45º
and
Examples
1. A particle is projected from a point O
with a speed of 
at an angle of
elevation of arcsin
.
a) Find the greatest height above O
reached by the particle .
b) The particle strikes the horizontal through O at Q find the distance OQ.
a) Using the maximum height
formula
You could also work out the answer by considering the vertical
component of the velocity and then use the constant acceleration equations.
Vertical component is given by
At maximum height velocity equals zero.
Using:
b) OQ is the horizontal range
Which can be expressed as:
2. A girl hits a ball at an
angle arctan
to the horizontal from a point O which is 0.5m
above level ground. The initial speed of the ball is 
. The ball just clears a fence which is a
horizontal distance of 18m from the girl. By modelling
the ball as a particle find the time taken for the ball to reach the fence and
the height of the fence.
Horizontal component of the velocity is Vcosα.
Therefore time in flight is
given by:
We must now consider the motion vertically to calculate the height of the
fence. The vertical component of the
velocity is:
and
by using
Fence
height = 12.9m
Questions A
1 A cricket ball is
hit from a point which is 0.9m above horizontal ground. It is given an initial
speed of 
at an angle
of elevation of 27 º.
Find:
a) the time taken for the ball to reach
the ground.
b) the horizontal distance covered by the
ball.
2 A tennis ball is
served from a height of 2.6m at horizontal speed of 
. The net is 0.9m high and 13m horizontally from the
server. Modelling
the ball as a particle, determine whether the ball clears the net and if so by
what distance.
3 A golfer hits a ball
with velocity 
at an angle
θ above the horizontal, where 
. Find the time for which the ball is at least 12m above the
ground.
4 A particle is
projected from a point O with speed 
at an angle 
above the horizontal.
Find
a) the time the particle takes to reach
the point P whose horizontal displacement is 96 metres,
b) the height of P above O,
c) the speed of the particle 2 seconds
after projection.
5 A cannon ball fired
at an angle of 10º has a range, on a horizontal plane, of 1.25km. Ignoring air
resistance, find the speed of projection.
6 A ball is projected
with velocity 
. If the range on the horizontal plane is 60m, find the two
possible angles of projection.
7 At time t seconds,
where t≥0, the velocity v ms-1
of a particle Q moving in a straight line is given by
When t = 0, Q is at point O.
a) Find the acceleration of Q at time t.
b) Calculate the time at which Q returns to O.
Differentiating and
Integrating Vectors
The constant
acceleration equations were used extensively in M1 but in M2 displacement is a
function of time. It follows then that
velocity, which is the rate of change of displacement, can be found by
differentiating the displacement function.
Displacement
Differentiate Velocity Integrate
(Remember
the constant)
Acceleration
If r is the position vector, and using dot
notation: -
Example 1
The position
vector of a particle Q at time t
is 
(r measured in metres).
Find the initial
position vector and show that the acceleration is constant
Initial position
is when t=0
Remembering that 
Since the
acceleration vector has no variable t it
is said to be constant.
Integrating Vectors
Since a vector has
both i and j components it will also
have two separate functions with respect to time.
Therefore if 
We need to
integrate each function separately, remembering the constants.
Therefore 
Where 
are constants of integration.
R
is found by integrating the velocity function with respect to time.
Example 2
A particle moves
such that at time t
At
time t = 0 the particle has a position vector 5i – 6j
Find
the position vector at time t.
The
position vector is found by integrating the velocity vector.
At
time t = 0, r = 5i – 6j 
,
Example
3
A
particle Q has position vector ( 25i – 40j )m
at time t = 0 relative to the origin. Q moves with constant acceleration and is
equal to 
. When t = 0,