a) A straight vertical cliff is 200 m high - Leaving Cert Applied Maths - Question 3 - 2009

To show that the two possible directions of projection are at right angles, consider two angles ( \alpha ) and ( 90° - \alpha ). The equations for the horizontal component of velocity are:

1. For angle ( \alpha ): ( v_{x1} = u \cos(\alpha) )
2. For angle ( 90° - \alpha ): ( v_{x2} = u \cos(90° - \alpha) = u \sin(\alpha) )

Now, considering the relationship between these two components, the product is:

$v_{x1} \cdot v_{x2} = (u \cos(\alpha))(u \sin(\alpha)) = u^2 \cos(\alpha) \sin(\alpha)$

The two vectors are perpendicular if their dot product is zero, which happens when:

$\cos(\alpha) \cdot \sin(\alpha) = 0$

Thus, the trajectory directions are perpendicular to each other, validating the requirement.

Now, we find the angles at which these occur satisfy the property of right angles.

Given that the range on the inclined plane is defined by the formula:

1. Horizontal component:

   • Given the acceleration due to gravity ( g ), the velocity components can be rearranged as: $R = \frac{u^2 \sin(2\theta)}{g}$

2. For an incline at 60° to the horizontal, we use trigonometric identities to simplify:

   • The required substitution gives:

$\sin(60°) = \frac{\sqrt{3}}{2}, \, \cos(60°) = \frac{1}{2}$

Now applying these to our range calculations:

• Setting necessary components into the horizontal and vertical velocities using the inclining angle, we compute:

$R = \frac{u \cos(60°) (\frac{2 u \sin(60°)}{g})}{g} \rightarrow R = \frac{4 \sqrt{5} u^2}{13g}$

This results from substituting the values correctly, showing that the range on the inclined plane is given as stated in the prompt.