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This approach retains the coordinates used in the normal case (in which projectile returns to the same horizontal level). In this consideration, the description of projectile motion is same as normal case except that motion is aborted in the mid-air by the incline. Had incline been not there, the projectile would have continued with its motion as shown in the figure.
When the projectile is allowed to return to the projection level, then the point of return is (OQ,0), where OQ is the horizontal range. This position of point of return changes to a new point (x,y), specified by the angle of elevation “α” of the wedge with respect to horizontal as shown in the figure.
From the triangle OPQ,
$$\mathrm{cos}\alpha =\frac{x}{OP}=\frac{x}{R}$$
The range of the projectile is given by :
$$\Rightarrow R=\frac{x}{\mathrm{cos}\alpha}$$
The strategy here is to determine “x” i.e. “OQ” considering the motion as normal projectile motion. Thus, we shall first determine “x” and then using above relation, we obtain the relation for the range of flight along the incline. Now, considering motion in horizontal direction , we have :
$$x=u\mathrm{cos}\theta XT$$
where “T” is the time of flight of projectile motion on the incline. It is given as determined earlier :
$$T=\frac{2u\mathrm{sin}\left(\theta -\alpha \right)}{g\mathrm{cos}\alpha}$$
Substituting in the epression of “x”, we have :
$$\Rightarrow x=\frac{u\mathrm{cos}\theta X2u\mathrm{sin}\left(\theta -\alpha \right)}{g\mathrm{cos}\alpha}$$
$$\Rightarrow x=\frac{{u}^{2}2\mathrm{sin}\left(\theta -\alpha \right)\mathrm{cos}\theta}{g\mathrm{cos}\alpha}$$
We simplify this relation, using trigonometric relation as given here :
$$\mathrm{sin}C-\mathrm{sin}D=2\mathrm{sin}\left(\frac{C-D}{2}\right)\mathrm{cos}\left(\frac{C+D}{2}\right)$$
Comparing right hand side of the equation with the expression in the numerator of the equation of “x”, we have :
$$C-D=2\theta -2\alpha $$
$$C+D=2\theta $$
Adding, we have :
$$\Rightarrow C=2\theta -\alpha $$
$$\Rightarrow D=\alpha $$
Thus, we can write :
$$\Rightarrow 2\mathrm{sin}\left(\theta -\alpha \right)\mathrm{cos}\theta =\mathrm{sin}\left(2\theta -\alpha \right)-\mathrm{sin}\alpha $$
Substituting the expression in the equation of “x”,
$$\Rightarrow x=\frac{{u}^{2}}{g\mathrm{cos}\alpha}\{\mathrm{sin}\left(2\theta -\alpha \right)-\mathrm{sin}\alpha \}$$
Using the relation connecting horizontal range “x” with the range on incline, “R”, we have :
$$R=\frac{x}{\mathrm{cos}\alpha}$$
$$\Rightarrow R=\frac{{u}^{2}}{g{\mathrm{cos}}^{2}\alpha}\{\mathrm{sin}\left(2\theta -\alpha \right)-\mathrm{sin}\alpha \}$$
Thus, we get the same expression for range as expected. Though the final expressions are same, but the understanding of two approaches is important as they have best fit application in specific situations.
A typical projection down the incline is shown with a reference system in which “x” and “y” axes are directions along incline and perpendicular to incline. The most important aspect of the analysis of this category of projectile motion is the emphasis that we put on the convention for measuring angles.
The angle of projection and angle of incline both are measured from a horizontal line. The expression for the time of flight is obtained by analyzing motion in vertical directions. Here we present the final results without working them out as the final forms of expressions are suggestive.
1: Components of initial velocity
$${u}_{x}=u\mathrm{cos}\left(\theta +\alpha \right)$$
$${u}_{y}=u\mathrm{sin}\left(\theta +\alpha \right)$$
2: Components of acceleration
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