Motion In A Plane
§§ PROJECTILE
Projectile is any body projected into the air at an angle othe than 90o with the horizontal near the surface of the earth. The path followed by a projectile is called its trajectory.
The following assumptions are made in the projectile motion
1. The acceleration due to gravity 'g' is constant over the range of motion.
2. The air resistance is negligible.
3. The ground on which the projectile thrown is refernce level.
§§ Equation of the Path (Trajectory)
Consider the motion of an object projected from the origin O of the co-ordinate system, with the initial velocity “u” inclined at an angle\(\alpha\) with the horizontal or the X-axis as shown in figure.
The motion of the object P can be studied by resolving its velocity “u” in the horizontal and the vertical directions as shown.
Horizontal component of the velocity is , \(
v_x = u\cos \alpha
\)
Vertical component of the velocity is vy = u sin\(\alpha\)
The horizontal component of velocity is vx. It shall remain constant as no acceleration is acting in the horizontal direction (a = 0)
The initial velocity in the vertical direction vy shall go on decreasing because of the constant deceleration due to gravity (a = – g).
The object, therefore, is having horizontal and vertical motions simultaneously.The resultant motion would be the vector sum of these two motions and the path followed would be
curvilinear.
Let P be the position of the object after a time “t” then distance travelled in the horizontal direction in time t, is x = (u cos a)t ......... (4.1)
The distance travelled in the vertical direction in time t is
y = (u sin a) t – \(
\frac{1}
{2}
\) gt2 ......... (4.2)
The equations (4.1) and (4.2) are the time displacement relations for the projectile.
Eliminating the time t, we can obtain a relationship between x and y or the equation of the path of the projectile.
From equation (4.1) we can write t = \(
\frac{x}
{{u\cos \alpha }}
\)
On substituting the value of “t” in equation (4.2) we can write
y = (u sin a)\(
\left( {\frac{x}
{{u\cos \alpha }}} \right)
\) –\(
\frac{1}
{2}
\)g \(
\left( {\frac{x}
{{u\cos \alpha }}} \right)^2
\)
y = (tan a) x – \(
\left( {\frac{g}
{{2u^2 \cos ^2 \alpha }}} \right)x^2
\) .......... (4.3)
The above equation is of the form y = Ax + Bx2 and represents a parabola. Thus the path of a projectile is a parabola.