Given initial velocity u = (20sqrt(3)i + 20j) ms^-1.

Here is the solution to question 3: 3) A particle is projected with initial velocity (20sqrt(3)i + 20j) ms^-1. Given initial velocity u = (20sqrt(3)i + 20j) ms^-1. This means the horizontal component of initial velocity is u_x = 20sqrt(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 a = -10j 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 = 20sqrt(3) ms^-1. The vertical velocity changes due to gravity: v_y = u_y - gt. At t=1 s: v_x = 20sqrt(3)\, ms^-1 v_y = 20 - (10)(1) = 10\, ms^-1 So, the velocity vector at t=1 s is v = (20sqrt(3)i + 10j) ms^-1. Step 2: Calculate the speed (magnitude of velocity). |v| = sqrt(v_x^2 + v_y^2) |v| = sqrt((203))^2 + (10)^2 |v| = sqrt((400 × 3) + 100) |v| = sqrt(1200 + 100) |v| = sqrt(1300) = sqrt(100 × 13) = 10sqrt(13)\, ms^-1 Step 3: Calculate the direction. The direction is given by the angle with the horizontal. = (v_y)/(v_x) = (10)/(20sqrt(3)) = (1)/(2sqrt(3)) = ((1)/(2sqrt(3))) ≈ 16.1^ The speed of the particle after 1 second is 10sqrt(13)\, ms^-1 and its direction is ((1)/(2sqrt(3))) 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 - (1)/(2)gt^2. Set y=0: 0 = u_y T - (1)/(2)gT^2 0 = T(u_y - (1)/(2)gT) This gives T=0 (initial projection) or u_y - (1)/(2)gT = 0. (1)/(2)gT = u_y T = (2u_y)/(g) Step 2: Substitute the values u_y = 20 ms^-1 and g = 10 ms^-2. T = (2 × 20)/(10) = (40)/(10) = 4\, s The time of flight of the particle is 4\, 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 = 20sqrt(3) ms^-1 and T = 4 s. R = (20sqrt(3))(4) = 80sqrt(3)\, m The range of the particle is 80sqrt(3)\, 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 = (400)/(20) = 20\, m The maximum height attained by the particle is 20\, 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 = (x)/(u_x) = (x)/(20sqrt(3)) Step 2: Substitute this expression for t into the equation for vertical displacement y. y = u_y t - (1)/(2)gt^2 y = (20)((x)/(20sqrt(3))) - (1)/(2)(10)((x)/(20sqrt(3)))^2 Step 3: Simplify the equation. y = (x)/(sqrt(3)) - 5((x^2)/(400 × 3)) y = (x)/(sqrt(3)) - 5((x^2)/(1200)) y = (x)/(sqrt(3)) - (x^2)/(240) The Cartesian equation of the path of the particle is