5.6 Projectile motion on an incline  (Page 3/7)

Range of flight

First thing that we should note that we do not call “horizontal range” as the range on the incline is no more horizontal. Rather we simply refer the displacement along x-axis as “range”. We can find range of flight by considering motion in both “x” and “y” directions. Note also that we needed the same approach even in the normal case. Let “R” be the range of projectile motion.

The motion along x-axis is no more uniform, but decelerated. This is the major difference with respect to normal case.

$$x=u_{x}T-\frac{1}{2}{a}_x{T}^2$$

Substituting value of “T” as obtained before, we have :

$$R=\frac{u\cos \left(\theta -\alpha \right)X2u\sin \left(\theta -\alpha \right)}{g\cos \alpha }-\frac{g\sin \alpha X4u^2\sin {}^2\left(\theta -\alpha \right)}{2g^2{\cos }^2\alpha }$$

$$\Rightarrow R=\frac{u^2}{g{\cos }^2\alpha }\{2\cos \left(\theta -\alpha \right)\sin \left(\theta -\alpha \right)\cos \alpha -\sin \alpha X2\sin {}^2\left(\theta -\alpha \right)\}$$

Using trigonometric relation, $2{\sin }^2\left(\theta -\alpha \right)=1-\cos 2\left(\theta -\alpha \right)$ ,

$$\Rightarrow R=\frac{u^2}{g\cos {}^2\alpha }[\sin 2\left(\theta -\alpha \right)\cos \alpha -\sin \alpha \{1-\cos 2\left(\theta -\alpha \right)\left\}\right]$$

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

We use the trigonometric relation, $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$ ,

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

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

This is the expression for the range of projectile on an incline. We can see that this expression reduces to the one for the normal case, when $\alpha =0$ ,

$$\Rightarrow R=\frac{u^2\sin 2\theta }{g}$$

Maximum range

The range of a projectile thrown up the incline is given as :

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

We see here that the angle of incline is constant. The range, therefore, is maximum for maximum value of “sin(2θ – α)”. Thus, range is maximum for the angle of projection as measured from horizontal direction, when :

$$\sin \left(2\theta -\alpha \right)=1$$

$$\Rightarrow \sin \left(2\theta -\alpha \right)=\sin \pi /2$$

$$\Rightarrow 2\theta -\alpha =\pi /2$$

$$\Rightarrow \theta =\pi /2+\alpha /2$$

The maximum range, therefore, is :

$$\Rightarrow R_{\max }=\frac{u^2}{g\cos {}^2\alpha }\left(1-\sin \alpha \right)$$