#### Answer

The ball was thrown from a height of 19.6 meters.

#### Work Step by Step

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.