5.6 Projectile motion on an incline  (Page 2/7)

It also follows that we measure other angles from the horizontal – even if we select x-coordinate in any other direction like along the incline. This convention avoids confusion. For example, the angle of incline “α” is measured from the horizontal. The horizontal reference, therefore, is actually a general reference for measurement of angles in the study of projectile motion.

Now, let us have a look at other characterizing aspects of new analysis set up :

1: The coordinate “x” is along the incline – not in the horizontal direction; and the coordinate “y” is perpendicular to incline – not in the vertical direction.

2: Angle with the incline

From the figure, it is clear that the angle that the velocity of projection makes with x-axis (i.e. incline) is “θ – α”.

3: The point of return

The point of return is specified by the coordinate R,0 in the coordinate system, where “R” is the range along the incline.

4: Components of initial velocity

$$u_x=u\cos \left(\theta -\alpha \right)$$

$$u_y=u\sin \left(\theta -\alpha \right)$$

5: The components of acceleration

In order to determine the components of acceleration in new coordinate directions, we need to know the angle between acceleration due to gravity and y-axis. We see that the direction of acceleration is perpendicular to the base of incline (i.e. horizontal) and y-axis is perpendicular to the incline.

Thus, the angle between acceleration due to gravity and y – axis is equal to the angle of incline i.e. “α”. Therefore, components of acceleration due to gravity are :

$$a_x=-g\sin \alpha$$

$$a_y=-g\cos \alpha$$

The negative signs precede the expression as two components are in the opposite directions to the positive directions of the coordinates.

6: Unlike in the normal case, the motion in x-direction i.e. along the incline is not uniform motion, but a decelerated motion. The velocity is in positive x-direction, whereas acceleration is in negative x-direction. As such, component of motion in x-direction is decelerated at a constant rate “gsin α”.

Time of flight

The time of flight (T) is obtained by analyzing motion in y-direction (which is no more vertical as in the normal case). The displacement in y-direction after the projectile has returned to the incline, however, is zero as in the normal case. Thus,

$$y=u_{y}T+\frac{1}{2}{a}_y{T}^2=0$$

$$\Rightarrow u\sin \left(\theta -\alpha \right)T+\frac{1}{2}\left(-g\cos \alpha \right)T^2=0$$

$$\Rightarrow T\{u\sin \left(\theta -\alpha \right)+\frac{1}{2}\left(-g\cos \alpha \right)T\}=0$$

Either,

$$T=0$$

or,

$$\Rightarrow T=\frac{2u\sin \left(\theta -\alpha \right)}{g\cos \alpha }$$

The first value represents the initial time of projection. Hence, second expression gives us the time of flight as required. We should note here that the expression of time of flight is alike normal case in a significant manner.

In the generic form, we can express the formula of the time of flight as :

$$T=\left|\frac{2u_y}{a_y}\right|$$

In the normal case, $u_y=u\sin \theta$ and $a_y=-g$ . Hence,

$$T=\frac{2u\sin \theta }{g}$$

In the case of projection on incline plane, $u_y=u\sin \left(\theta -\alpha \right)$ and $a_y=-g\cos \alpha$ . Hence,

$$T=\frac{2u\sin \left(\theta -\alpha \right)}{g\cos \alpha }$$

This comparison and understanding of generic form of the expression for time of flight helps us write the formula accurately in both cases.