Consider the motion of a particle which is projected in a direction making
an angle with the horizon. When we neglect drag, the only force
which acts upon the particle is its weight, (Fig. 66).
\begin{figure}
\includegraphics[scale=.8]{Fig66.eps}
\end{figure}
Taking the plane of motion to be the xy-plane, and applying Newton's laws of motion gives us the equations
{\mathbf x-axis} \\
{\mathbf y-axis} \\
where and are the components of the
acceleration along the x and y axes. Integrating equations (1) and (2) we get
Therefore the component of the velocity along the x-axis remains constant, while the
component along the y-axis changes uniformly. Let be the initial velocity of
the projection, then when , and .
Making these substitutions in the last two equations we obtain
Therefore
Then the total velocity at any instant is
and makes an angle with the horizon defined by
Integrating equations (3) and (4) we obtain
But when , , therefore , and consequently
It is interesting to note that the motions in the two directions are independent. The
gravitational acceleration does not affect the constant velocity along the x-axis, while
the motion along the y-axis is the same as if the body were dropped vertcally with
an initial velocity .
{bf The Path} - The equation of the path may be obtained by eliminating between equations
(7) and (8). This gives
which is the equation of a parabola.
{bf The Time of Flight} - When the projectile strikes the ground its y-coordinate is zero.
Therefore substituting zero for in equation (8) we get for the time of flight
{\mathbf The Range} - The range, or the total horizontal distance covered by the projectile,
is found by replacing in equation (7) by the value of in equation (10), or
by letting in equation (9). By either method we obtain
Note that a basic trigonometric identity was used to simpilfy the above equation.
Since and are constants the value of depends upon . It is evident from
equation (11) that is maximum when , or when . The
maximum range is, therefore,
In actual practice the angle of elevation which gives the maximum range is smaller
on account of the resistance of the air.
{\mathbf The Highest Point} - At the highest point . Therefore substituting
this value of in equation (4) we obtain or
for the time taken to reach the highest point. Subsituting this value of the time
in equation (8) we get for the maximum elevation
{\mathbf The Range for a Sloping Ground} - Let be the angle which the ground
makes with the horizon. Then the range is the distance , Fig. 67, where
is the point where the projectile strikes the sloping ground. The equation of the
line is
\begin {equation}
y = x \tan \beta
</math>
\begin{figure}
\includegraphics[scale=.8]{Fig67.eps}
\end{figure}
Eliminating between equations (14) and (9) we obtain the x-coordinate of the
point,
But , where .
Therefore
Thus for a given value of , is maximum when ,
that is, when .
When equations (15) and (16) reduce to equations (12) and (13), as they should.