Example 6-19.
A cannon ball of mass mis fired from a cannon with
an initial velocity v 0 inclined at an angle θwith the
horizontal as illustrated. Neglect air resistance and
find the equations of motion, maximum height, and
range of the cannon ball.
Solution: Let y=y(t)denote the vertical height at any time tand let x=x(t)denote
the horizontal distance at any time t. Consider the cannon ball at a position (x, y )
and examine the forces acting on it. In the y-direction the force due to the weight of
the cannon ball is W=mg, ( g= 32 ft/sec^2 ). The equation of motion in the y-direction
is represented as
md(^2) y
dt^2
=−W=−mg. (6 .50)
Forces in the x-direction like air resistance are neglected. Newton’s second law can
then be expressed
md(^2) x
dt^2
= 0. (6 .51)
Make note of the fact that whenever time t is the independent variable, the dot
notation
x ̇=dx
dt, x ̈=d(^2) x
dt^2
, y ̇=dy
dt
, y ̈=d
(^2) y
dt^2
(6 .52)
is often employed to denote derivatives. Using the dot notation the equations (6.50)
and (6.51) would be represented
̈y=−g and x ̈= 0 (6 .53)
Calculating the xand y-components of the initial velocity, the equations (6.50)
and (6.51) are solved subject to the initial conditions:
x(0) = 0, y(0) = 0
x ̇(0) = v 0 cos θ, y ̇(0) = v 0 sin θ,where v 0 is the initial speed and θis the angle of inclination of the cannon. Solving
the differential equations (6.50) and (6.51) by successive integrations gives
y ̇=−gt +c 1 ,y=−gt^2
2 +c^1 t+c^2 ,x ̇=c 3
x=c 3 t+c 4