A particle is projected from the top of a straight vertical cliff of height 25 m with velocity 15i + 20j - Leaving Cert Applied Maths - Question 3 - 2013

The maximum height above ground is calculated using: $s_y = u_y t + \frac{1}{2} a t^2$ Substituting $u_y = 20 \, \text{ms}^{-1}$, $t = 2 \, \text{s}$, and $a = -10 \, \text{ms}^{-2}$:

$s_y = 25 + (20)(2) + \frac{1}{2}(-10)(2^2)$

Calculating: $s_y = 25 + 40 - 20 = 45 \text{ m}$

For the entire flight time, we use: $s_y = u_y t + \frac{1}{2} a t^2$ Setting the vertical displacement from point A to the ground (25 m down), we have:

$-25 = 20t - 5t^2$ Rearranging to: $5t^2 - 20t - 25 = 0$ Using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 5$, $b = -20$, and $c = -25$:

$t = \frac{20 \pm \sqrt{(-20)^2 - 4(5)(-25)}}{2(5)}$ Calculating: $t = \frac{20 \pm \sqrt{400 + 500}}{10} = \frac{20 \pm 30}{10}$ Thus, $t = 5 \, \text{s}$.

The horizontal distance |AB| can be calculated with: $|AB| = u_x t$ Given that $u_x = 15 \, \text{ms}^{-1}$ and $t = 5 \, \text{s}$:

$|AB| = 15 \times 5 = 75 \text{ m}.$

The final velocities can be calculated:

• The horizontal velocity remains constant: $v_x = 15 \, \text{ms}^{-1}$.
• The vertical velocity can be calculated as: $v_y = u_y + at = 20 - 10 \times 5 = -30 \, \text{ms}^{-1}.$

Using Pythagoras’ theorem for total speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(15)^2 + (-30)^2} = \sqrt{225 + 900} = \sqrt{1125} = 15\sqrt{5} \approx 33.54 \, \text{ms}^{-1}.$