where c 1 , c 2 , c 3 , c 4 are constants of integration. The solution satisfying the initial
conditions can be expressed as
y=y(t) = −g
2t^2 + (v 0 sin θ)t
x=x(t) = (v 0 cos θ)t.(6 .54)These are parametric equations describing the position of the cannon ball. The
position vector describing the path of the cannon ball is given by
r =r (t) = (v 0 cos θ)tˆe 1 + (−g 2 t^2 + (v 0 sin θ)t)ˆe 2The maximum height occurs where the derivative dydt is zero, and the maximum range
occurs when the height yreturns to zero at some time t > 0 .The derivative dydt is zero
when thas the value t 1 =v 0 sin θ/g, and at this time,
ymax =y(t 1 ) = v(^20) sin^2 θ
2 g
, x =x(t 1 ) = v
(^20) sin2 θ
2 g
(6 .55)
The maximum range occurs when y= 0 at time t 2 = 2
v 0 sin θg ,and at this time,
xmax =x(t 2 ) = v
02 sin2 θ
g.Eliminating tfrom the parametric equations (6.54), demonstrates that the trajectory
of the cannon ball is a parabola.
Example 6-20. (Circular motion)
Consider a particle moving on a circle of radius rwith a constant angular velocity
ω= dθdt. Construct a cartesian set of axes with origin at the center of the circle.
Assume the position of the particle at any given time tis given by the position
vector
r =r (t) = rcos ω t ˆe 1 +rsin ω t ˆe 2 rand ωare constants.
The displacement of the particle as it moves around the circle is given by s=rθ and
the speed of the particle is
ds
dt =v=rdθdt =rω. The velocity of the particle is a vector
quantity given by
v =dr
dt=−rω sin ω t ˆe 1 +rω cos ω t ˆe 2 (6 .56)