A particle is projected with speed 40 m s$^{-1}$ from a point A on the top of a vertical cliff of height 30 m - Leaving Cert Applied Maths - Question 3 - 2019

The horizontal range can be found using the formula:

$R = u \cos \alpha \cdot t$

Substituting the values we found earlier:

• Using $\alpha = 22.5^{\circ}$ and $t = 4.49$ s:

$R = 40 \cos(22.5^{\circ}) \cdot 4.49 \approx 165.9 \, m$

To calculate the final speed at point C, we can use the formula:

$|v| = \sqrt{(v_{x})^2 + (v_{y})^2}$

Where:

• $v_{x} = 40 \cos(22.5^{\circ}) t$
• $v_{y} = 40 \sin(22.5^{\circ}) - gt$

Calculating:

1. $v_{y}$ at C is:

$v_{y} = 36.955 - 28.695 = 8.26 \, m s^{-1}$

2. $v_{x}$ is approximately:

$v_{x} \approx 36.79 \, m s^{-1}$

Therefore, the magnitude of the final speed is:

$|v| \approx 46.79 \, m s^{-1}$

To determine the time taken for the particle to reach point Q, we need to solve for $t$:

Using the formula for vertical motion:

$y_P - y_Q = v_{i_y}t - \frac{1}{2}gt^2$

With given lengths and speeds, substitute:

$1 = 20 \sin(60^{\circ}) t - \frac{1}{2} g t^2$

Rearranging and solving, we find:

$t \approx 3.057 \, s$

To find the speed of the particle at the point where it is travelling parallel to the inclined plane, we examine the vertical and horizontal components of motion:

1. The horizontal component of the velocity:

$v_{x} = 20 \sin(45^{\circ}) - g \cos(15^{\circ}) t$ 2. The vertical component:

$v_{y} = 20 \cos(45^{\circ}) \cdot t$

The final speed there is given by:

$v = \sqrt{v_{x}^{2} + v_{y}^{2}}\quad \text{where } v_{y} = 10.36 \, m s^{-1}$