A ball is thrown upwards with an initial velocity of 20 m/s. Ignoring air resistance, how high does it go? (Use \( g = 9.8 \, \text{m/s}^2 \))

["How High Does a Ball Go When Thrown Upwards with Initial Velocity of 20 m/s?", "When a ball is thrown upward with an initial velocity, its maximum height depends on the conversion of kinetic energy into potential energy. Ignoring air resistance simplifies the physics, allowing us to use fundamental equations of motion.", "Given:\n- Initial velocity, ( v_0 = 20 , \ ext{m/s} )\n- Acceleration due to gravity, ( g = 9.8 , \ ext{m/s}^2 ) (acting downward)\n- Final velocity at maximum height, ( v = 0 , \ ext{m/s} ) (momentarily stops)", "To find the maximum height ( h ), use the kinematic equation:", "[\nv^2 = v_0^2 - 2gh\n]", "At the peak, ( v = 0 ), so:", "[\n0 = v_0^2 - 2gh\n]", "Solve for ( h ):", "[\nh = \frac{v_0^2}{2g} = \frac{20^2}{2 \ imes 9.8} = \frac{400}{19.6} \approx 20.41 , \ ext{meters}\n]", "### Conclusion\nIgnoring air resistance, a ball thrown upward with an initial velocity of 20 m/s reaches a maximum height of approximately 20.41 meters. This calculation relies on basic principles of energy conservation and motion under constant acceleration.", "Understanding this concept is fundamental in physics and helps in sports, engineering, and everyday predictions of projectile motion."]