This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here's the solution for question 20: Given: Mass of the ball, m = 0.1\,kg Initial velocity, u = 80\,m/s Acceleration due to gravity, g = 9.8\,m/s^2 Part (iii): Kinetic energy as it leaves the ground. The kinetic energy as it leaves the ground is its initial kinetic energy. The formula for kinetic energy is KE = (1)/(2)mv^2. KE_initial = (1)/(2)mu^2 KE_initial = (1)/(2)(0.1\,kg)(80\,m/s)^2 KE_initial = (1)/(2)(0.1)(6400) KE_initial = 0.05 × 6400 KE_initial = 320\,J The kinetic energy as it leaves the ground is 320 J. Step 1: Calculate the maximum height (H_max). At the maximum height, the final velocity (v) of the ball is 0\,m/s. Using the kinematic equation v^2 = u^2 + 2as, where a = -g (since gravity acts downwards) and s = H_max: 0^2 = u^2 + 2(-g)H_max 0 = (80\,m/s)^2 - 2(9.8\,m/s^2)H_max 0 = 6400 - 19.6 H_max 19.6 H_max = 6400 H_max = (6400)/(19.6) H_max ≈ 326.53\,m Part (ii): Potential energy at its maximum height. The formula for potential energy is PE = mgh. PE_max = mgH_max PE_max = (0.1\,kg)(9.8\,m/s^2)(326.53\,m) PE_max = 320\,J The potential energy at its maximum height is 320 J. Part (i): Potential energy half way up. Half way up, the height is H_half = H_max2. H_half = (326.53\,m)/(2) H_half ≈ 163.265\,m Now, calculate the potential energy at this height: PE_half = mgH_half PE_half = (0.1\,kg)(9.8\,m/s^2)(163.265\,m) PE_half = 160\,J The potential energy half way up is 160 J.