The problem describes a technician sliding from a tower of height 80 m. We are given the acceleration due to gravity g = 10 N/Kg. We will assume the technician starts from rest and

Hey phirinathan906, good to see you again. The problem describes a technician sliding from a tower of height 80 m. We are given the acceleration due to gravity g = 10 N/Kg. We will assume the technician starts from rest and the question asks for the speed just before hitting the ground. Given: • Initial velocity, u = 0 m/s (assuming the technician starts from rest) • Height, h = 80 m • Acceleration due to gravity, g = 10 N/Kg = 10 m/s^2 We need to find the final velocity, v. Step 1: Identify the appropriate kinematic equation. To find the final velocity when initial velocity, acceleration, and displacement are known, we use the following equation: v^2 = u^2 + 2gh Step 2: Substitute the given values into the equation. v^2 = (0 m/s)^2 + 2(10 m/s^2)(80 m) Step 3: Calculate the square of the final velocity. v^2 = 0 + 1600 m^2/s^2 v^2 = 1600 m^2/s^2 Step 4: Calculate the final velocity by taking the square root. v = sqrt(1600 m)^2/s^2 v = 40 m/s The speed of the technician just before hitting the ground is 40 m/s. Send me the next one 📸