A particle is projected from point P, as shown in the diagram, with initial speed 40√3 m s⁻¹ at an angle of 60° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2017

To find the velocity after 4 seconds, we include the effects of gravitational acceleration. The horizontal velocity remains constant:

• The horizontal component remains the same: $v_x = 20\sqrt{3}$

• The vertical component changes due to gravity: $v_y = u_y - g t = 60 - 10 \times 4 = 20$

Thus, after 4 seconds, the velocity vector is: $\mathbf{v} = 20\sqrt{3} \mathbf{i} + 20 \mathbf{j}$

The greatest height is reached when the vertical velocity component is 0. Set: $v_y = u_y - gt = 0$ From this, we find time to reach the maximum height: $t = \frac{u_y}{g} = \frac{60}{10} = 6 \text{ seconds}$

Using this time to calculate the height: $h = u_y t - \frac{1}{2} g t^2 = 60(6) - \frac{1}{2}(10)(6^2)$
Calculating gives: $h = 360 - 180 = 180 \text{ m}$

Using the equation of motion for the vertical distance: $s = ut + \frac{1}{2} a t^2$ We know the particle lands at height 135 m, thus: $s = 135, u = 60, a = -10$ Now using the known values: $135 = 60t - \frac{1}{2}(10)t^2$ This simplifies to: $0 = -5t^2 + 60t - 135$ Using the quadratic formula: $t = \frac{60 \pm \sqrt{60^2 - 4(-5)(-135)}}{2(-5)}$ The two times solve to: $t = 9 \text{ seconds and } t = 3 \text{ seconds}$

The time above the cliff can be determined by solving the equations for both time intervals calculated earlier. Find while at the height of the cliff. Since the height function results in a parabolic trajectory, we need to calculate:

• Time from the lowest point above Q to the point reaching the same height again: The time taken for the particle to go up and come back down can be analyzed using: $t = \frac{u_y + \sqrt{u_y^2 - 2gh}}{g}$ where $h$ is vertical distance corresponding to 135 m. After calculations, the two segments are:
• Total time: t_{ ext{total}} = 9 \text{ seconds (total time above cliff) - Time at landing}}