Here is the solution to question 3: 3) A particle is projected with initial velocity $(20\sqrt{3}\mathbf{i} + 20\mathbf{j})$ ms$^{-1}$. Given initial velocity $\mathbf{u} = (20\sqrt{3}\mathbf{i} + 20\mathbf{j})$ ms$^{-1}$. This means the horizontal component of initial velocity is $u_x = 20\sqrt{3}$ ms$^{-1}$ and the vertical component is $u_y = 20$ ms$^{-1}$. We assume the acceleration due to gravity $g = 10$ ms$^{-2}$ acts downwards, so $\mathbf{a} = -10\mathbf{j}$ ms$^{-2}$. a) Find the speed and direction of the particle after 1 second. Step 1: Find the velocity components at time $t$. The horizontal velocity remains constant: $v_x = u_x = 20\sqrt{3}$ ms$^{-1}$. The vertical velocity changes due to gravity: $v_y = u_y - gt$. At $t=1$ s: $$ v_x = 20\sqrt{3}\, \text{ms}^{-1} $$ $$ v_y = 20 - (10)(1) = 10\, \text{ms}^{-1} $$ So, the velocity vector at $t=1$ s is $\mathbf{v} = (20\sqrt{3}\mathbf{i} + 10\mathbf{j})$ ms$^{-1}$. Step 2: Calculate the speed (magnitude of velocity). $$ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} $$ $$ |\mathbf{v}| = \sqrt{(20\sqrt{3})^2 + (10)^2} $$ $$ |\mathbf{v}| = \sqrt{(400 \times 3) + 100} $$ $$ |\mathbf{v}| = \sqrt{1200 + 100} $$ $$ |\mathbf{v}| = \sqrt{1300} = \sqrt{100 \times 13} = 10\sqrt{13}\, \text{ms}^{-1} $$ Step 3: Calculate the direction. The direction is given by the angle $\alpha$ with the horizontal. $$ \tan \alpha = \frac{v_y}{v_x} = \frac{10}{20\sqrt{3}} = \frac{1}{2\sqrt{3}} $$ $$ \alpha = \arctan\left(\frac{1}{2\sqrt{3}}\right) \approx 16.1^\circ $$ The speed of the particle after 1 second is $\boxed{10\sqrt{13}\, \text{ms}^{-1}}$ and its direction is $\boxed{\arctan\left(\frac{1}{2\sqrt{3}}\right) \text{ above the horizontal}}$. b) The time of flight of the particle. Step 1: The time of flight $T$ is the time when the vertical displacement $y$ is zero. The vertical displacement is given by $y = u_y t - \frac{1}{2}gt^2$. Set $y=0$: $$ 0 = u_y T - \frac{1}{2}gT^2 $$ $$ 0 = T\left(u_y - \frac{1}{2}gT\right) $$ This gives $T=0$ (initial projection) or $u_y - \frac{1}{2}gT = 0$. $$ \frac{1}{2}gT = u_y $$ $$ T = \frac{2u_y}{g} $$ Step 2: Substitute the values $u_y = 20$ ms$^{-1}$ and $g = 10$ ms$^{-2}$. $$ T = \frac{2 \times 20}{10} = \frac{40}{10} = 4\, \text{s} $$ The time of flight of the particle is $\boxed{4\, \text{s}}$. c) The range of the particle. Step 1: The range $R$ is the horizontal distance covered during the time of flight. $$ R = u_x T $$ Step 2: Substitute the values $u_x = 20\sqrt{3}$ ms$^{-1}$ and $T = 4$ s. $$ R = (20\sqrt{3})(4) = 80\sqrt{3}\, \text{m} $$ The range of the particle is $\boxed{80\sqrt{3}\, \text{m}}$. d) The maximum height attained by the particle. Step 1: At maximum height, the vertical component of velocity $v_y = 0$. We can use the kinematic equation $v_y^2 = u_y^2 - 2gH$, where $H$ is the maximum height. $$ 0^2 = (20)^2 - 2(10)H $$ $$ 0 = 400 - 20H $$ Step 2: Solve for $H$. $$ 20H = 400 $$ $$ H = \frac{400}{20} = 20\, \text{m} $$ The maximum height attained by the particle is $\boxed{20\, \text{m}}$. e) The Cartesian equation of the path of the particle. Step 1: Express time $t$ in terms of horizontal displacement $x$. The horizontal displacement is $x = u_x t$. $$ t = \frac{x}{u_x} = \frac{x}{20\sqrt{3}} $$ Step 2: Substitute this expression for $t$ into the equation for vertical displacement $y$. $$ y = u_y t - \frac{1}{2}gt^2 $$ $$ y = (20)\left(\frac{x}{20\sqrt{3}}\right) - \frac{1}{2}(10)\left(\frac{x}{20\sqrt{3}}\right)^2 $$ Step 3: Simplify the equation. $$ y = \frac{x}{\sqrt{3}} - 5\left(\frac{x^2}{400 \times 3}\right) $$ $$ y = \frac{x}{\sqrt{3}} - 5\left(\frac{x^2}{1200}\right) $$ $$ y = \frac{x}{\sqrt{3}} - \frac{x^2}{240} $$ The Cartesian equation of the path of the particle is