5.6 Projectile motion on an incline  (Page 6/7)

This situation can be handled with a reoriented coordinate system as shown in the figure. Here, angle of projection with respect to x - direction is (θ-α) and acceleration in y - direction is "g cosα". Now, total time of flight for projectile motion, when points of projection and return are on same level, is :

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

Replacing "θ" by "(θ-α)" and "g" by "gcosα", we have formula of time of flight over the incline :

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

Now, θ = 60° , α = 30°, u = 10 m/s. Putting these values,

$$\Rightarrow T=\frac{2X10\sin \left({60}^0-{30}^0\right)}{g\cos {30}^0}=\frac{20\sin {30}^0}{10\cos {30}^0}=\frac{2}{\sqrt{3}}$$

Hence, option (d) is correct.

We have discussed that projectile motion on an incline surface can be rendered equivalent to projectile motion on plane surface by reorienting coordinate system as shown here :

In this reoriented coordinate system, we need to consider component of acceleration due to gravity along y-direction. Now, time of flight is given by :

$$t=\frac{2u_y}{g\cos \theta }$$

Let us first consider the projectile thrown from point "O". Considering the angle the velocity vector makes with the horizontal, the time of flight is given as :

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

For the projectile thrown from point "A", the angle with horizontal is zero. Hence, the time of flight is :

$$\Rightarrow t_A=\frac{2u\sin \left(2X0+\theta \right)}{g\cos \theta }=\frac{2u\tan \theta }{g}$$

Thus, we see that times of flight in the two cases are equal.

$$\Rightarrow t_A=t_O$$

Hence, option (a) is correct.

The velocity in y-direction can be determined making use of the fact that a ball under constant acceleration like gravity returns to the ground with the same speed, but inverted direction. The component of velocity in y-direction at the end of the journey, in this case, is :

$$v_y=u_y=-20\sin {30}^0=-20X\frac{1}{2}=-10m/s$$

$$\left|v_y\right|=10m/s$$

Hence, option (a) is correct.

We can interpret the equation obtained for the range of projectile :

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

The range is maximum for the maximum value of "sin(2θ-α) “ :

$$\sin \left(2\theta -\alpha \right)=1=\sin {90}^0$$ $$\Rightarrow 2\theta -{30}^0={90}^0$$ $$\Rightarrow \theta ={60}^0$$

Hence, option (b) is correct.