## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We first find the time $t$ of the ball's flight; $t = \frac{x}{v_x} = \frac{50~m}{25~m/s}$ $t = 2.0~s$ We then find the initial height of the ball; $y = \frac{1}{2}gt^2$ $y = \frac{1}{2}(9.8~m/s^2)(2.0~s)^2$ $y = 19.6~m$ The ball was thrown from a height of 19.6 meters.