1.6 Projectile motion on an incline  (Page 5/7)

Problem : A ball is projected in horizontal direction from an elevated point “O” of an incline of angle “α” with a speed “u”. If the ball hits the incline surface at a point “P” down the incline, find the coordinates of point “P”.

Solution : We can answer this question, using the relation of coordinates with range “R” as :

$$x=R\cos \alpha$$

$$y=-R\sin \alpha$$

Now, range of the flight for the downward flight is given as :

$$R=\frac{u^2}{g\cos {}^2\alpha }\{\sin \left(2\theta +\alpha \right)+\sin \alpha \}$$

The important thing to realize here is that the ball is projected in horizontal direction. As we measure angle from the horizontal line, it is evident that the angle of projection is zero. Hence,

$$\theta =0^0$$

Putting in the equation for the range of flight, we have :

$$R=\frac{2u^2\sin \alpha }{g{\cos }^2\alpha }$$

Therefore, coordinates of the point of return, “P”, is :

$$\Rightarrow x=R\cos \alpha =\frac{2u^2\sin \alpha }{g\cos \alpha }$$

$$\Rightarrow x=\frac{2u^2\tan \alpha }{g}$$

Similarly,

$$\Rightarrow y=-R\sin \alpha =-\frac{2u^2\sin \alpha }{g{\cos }^2\alpha }\sin \alpha$$

$$\Rightarrow y=-\frac{2u^2{\tan }^2\alpha }{g}$$

Problem : A ball is projected in horizontal direction from an elevated point “O” of an incline of angle “α” with a speed “u”. If the ball hits the incline surface at a point “P” down the incline, find the coordinates of point “P”.

Solution : We draw or shift the horizontal base and represent the same by the line QP as shown in the figure below.

From the general consideration of projectile motion, the vertical displacement, “y”, is :

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

$$\Rightarrow y=0-\frac{1}{2}g{T}^2$$

Considering the magnitude of vertical displacement only, we have :

$$\Rightarrow y=\frac{1}{2}g{T}^2$$

On the other hand, consideration of motion in x-direction yields,

$$x=u_{x}T=uT$$

Since, we aim to find the coordinates of point of return “P”, we eliminate “T” from the two equations. This gives us :

$$y=\frac{g{x}^2}{2u^2}$$

From the triangle OPQ, we have :

$$\tan \alpha =\frac{y}{x}$$

$$\Rightarrow y=x\tan \alpha$$

Combining two equations, we have :

$$y=x\tan \alpha =\frac{g{x}^2}{2u^2}$$

$$\Rightarrow \tan \alpha =\frac{gx}{2u^2}$$

$$\Rightarrow x=\frac{2u^2\tan \alpha }{g}$$

and

$$\Rightarrow y=x\tan \alpha =\frac{2u^2\tan {}^2\alpha }{g}$$

The y-coordinate, however, is below origin and is negative. Thus, we put a negative sign before the expression :

$$\Rightarrow y=-\frac{2u^2\tan {}^2\alpha }{g}$$