Question

 A boy on a bridge throws a stone horizontally with a speed of 25 m/s releasing the stone from a point 19.6 m above the surface of the river. How far from a point directly below the boy will the stone strikes the water?

Answer 1

Vertical and horizontal motions are independent. Since there is no forces acting on the stone in horizontal direction, its velocity in this direction does not change $v_h = v_{0h} = 25m/s$.

Time of its fall in vertical direction can be found from the kinematic equation:

$$h =\dfrac{gt^2}{2}$$

where $h = 19.6m$ is the height of the surface, $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtian:

$$t = \sqrt{\dfrac{2h}{g}}$$

In this time the stone covers the following horizontal distance:

$$d = tv_h =v_h \sqrt{\dfrac{2h}{g}}\\ d = 25m/s \sqrt{\dfrac{2\cdot 19.6m}{9.81m/s^2}} \approx 50m$$

Answer. 50 m.