A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise ?

At the highest point, the velocity of the ball is 0 m/s. When the ball reaches half its maximum height, it has a speed of 10 m/s. Let's denote the maximum height as h. At the point where the ball reaches half its maximum height (h/2), the velocity is 10 m/s. Using the equation for uniformly accelerated motion, v^2 = u^2 + 2as, where v is the final velocity (10 m/s), u is the initial velocity (0 m/s at the highest point), a is the acceleration (-g = -9.8 m/s^2, since it's accelerating downward), and s is the displacement (h/2), we can solve for h. The equation becomes 10^2 = 0^2 + 2 (-9.8) (h/2), which simplifies to 100 = -9.8h. Solving for h gives h = -100 / 9.8 = 10.2 m. However, this calculation seems to have been approached with an incorrect application of the equation. The correct approach involves recognizing that the ball's kinetic energy at the half-height point is converted from its potential energy at the maximum height. The potential energy at the maximum height (mgh) is converted into kinetic energy (0.5mv^2) at the half-height point plus the potential energy at the half-height point (mg(h/2)). The equation to solve should relate the initial potential energy at the maximum height to the sum of the kinetic energy and the potential energy at the half-height point. The correct formula should account for the conversion of potential to kinetic energy and the relationship between the height and the velocity at the half-height point.