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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20. A ball of mass 0.1 kg is thrown vertically upwards with an initial velocity of 80 m/s. Calculate the potential energy (i) half way up (ii) at its maximum height. What is its kinetic energy as it leaves the ground? Given: Mass m = 0.1 kg Initial velocity v_0 = 80 m/s Acceleration due to gravity g = 9.8 m/s^2 (assumed) Step 1: Calculate the maximum height (H_max). At maximum height, the final velocity v_f = 0 m/s. Using the kinematic equation v_f^2 = v_0^2 + 2ad: 0^2 = v_0^2 - 2gH_max H_max = (v_0^2)/(2g) Substitute the given values: H_max = (80 m/s)^22 × 9.8 m/s^2 = 6400 m^2/s^219.6 m/s^2 H_max ≈ 326.53 m Step 2: Calculate the potential energy at its maximum height. Potential energy PE = mgh. PE_max = m g H_max PE_max = (0.1 kg)(9.8 m/s^2)(326.53 m) PE_max = 320 J Alternatively, by conservation of energy, the potential energy at maximum height is equal to the initial kinetic energy: PE_max = (1)/(2) m v_0^2 = (1)/(2) (0.1 kg)(80 m/s)^2 = (1)/(2) (0.1)(6400) = 320 J The potential energy at its maximum height is 320 J. Step 3: Calculate the potential energy half way up. The height half way up is H_half = H_max2. H_half = 326.53 m2 = 163.265 m PE_half = m g H_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. Step 4: Calculate its kinetic energy as it leaves the ground. As the ball leaves the ground, its velocity is its initial velocity v_0. Kinetic energy KE = (1)/(2) m v^2. KE_ground = (1)/(2) m v_0^2 KE_ground = (1)/(2) (0.1 kg)(80 m/s)^2 KE_ground = (1)/(2) (0.1)(6400) KE_ground = 320 J The kinetic energy as it leaves the ground is 320 J.