3. (a) A particle is projected from a point P on horizontal ground - Leaving Cert Applied Maths - Question 3 - 2011

For the second angle of projection, we apply the following:

Equating the vertical position gives: $35 ext{ sin}( heta) imes t - 4.9t^2 = 50$ And for the horizontal: $35 ext{ cos}( heta) imes t = 50$

By utilizing these two equations, we can find the second angle. Setting up the equations: $50 - 10 an( heta) = -4.9t^2$ Using the relationship between angles, we can find: $an( heta - 2) = (2 an( heta - 3))$

This gives us a second possible angle of projection: $heta = 71.6°$

When a particle is projected down a plane, the components of the velocity must be equated according to the angle:

The horizontal speed component: $v_x = 10 ext{ cos}(45°) + 5 ext{ sin}( heta)$ The vertical component yields: v_y = 10 ext{ sin}(45°) - grac{10 ext{ cos}( heta)}{g ext{ cos}( heta)}

Using the equations of motion and the given landing angle, we can apply: $an(\theta) = \frac{-v_y}{v_x}$

We determine: $\tan(\theta) = \frac{1}{4}$ Solving for $\alpha$ provides: $\alpha = 56.3°$

We know that: $\text{Rebound velocity} = e \times \text{Initial velocity}$ Given: $v_r = 20v_i$ Thus, setting: $\sqrt{\frac{20}{2}} + \sqrt{5e\sqrt{2}} = e$

Solving this results in the coefficient of restitution: $e = \frac{1}{\sqrt{2}}$