Please have a read of the problem description:


How can I find the relationship between angle theta and phi???

Also, by considering energy (or otherwise), how can I calculate the vertical height reached when ball ceases bouncing, for u=100m/s?

Here's my working so far:


Can someone check if the angles can be related like so?

Thank you so much!!


1 Answer 1


Decompose the initial velocity into its horizontal and vertical components $(u_H, u_V)=(u \cos \theta, u \sin \theta)$. The ball reaches the top of the parabola exactly when the vertical component of velocity is $0$.

At that point, energy is

$$mgh = \frac{1}{2} m u_H^2$$ since $u_H$ is constant. So you get that the ball meets the plane at a height $h= \frac{u_H^2}{2g}$.

Once you find the time $T$ in which the ball meets the plane, you can conclude that $$\sin \phi = \frac{h}{Tu_H}$$ which gives you the relation between $\theta$ and $\phi$.

This time can be found with the formula $$h = u_V T-\frac{1}{2}gT^2$$

so that $$T=\frac{1}{g} (u_V - \sqrt{u_V^2 - 8gh}) = \frac{1}{g} (u_V - \sqrt{u_V^2 - 4u_H^2})$$ hence $$\sin \phi = \frac{1}{u \cos \theta}\frac{h}{T}= \frac{1}{u \cos \theta} (u\sin \theta - \frac{1}{2}g T) =$$ $$= \frac{\sqrt{\sin^2 \theta - 4 \cos^2 \theta}}{\cos \theta}$$

  • $\begingroup$ thanks! I see now...does the ball stop bouncing when theta= phi i.e. when the angle of projection equals the angle of incline so the ball technically "rolls forward" since it doesn't leave the surface? $\endgroup$ Mar 7, 2015 at 9:37

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