Question

A stone is thrown horizontally off a 25 m high building. It hits the ground a distance of 20 m from the foot of the building. What is the initial velocity of the stone?

Answer 1

Since the velocity in horizontal direction does not change during the motion, the initial velocity can be found as follows:

$$v_0 = \dfrac{d}{t}$$

where $d = 20 m$ is the distance travelled in horizontal direction, and $t$ is the time of fall.

From the kinematic equation

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

where $h=25m$ is the distance travelled in vertical direction, one can express the time of fall:

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

Substituting it into the expression for velocity, obtain:

$$v_0 = d\sqrt{\dfrac{g}{2h}} =20m\cdot \sqrt{\dfrac{9.8m/s^2}{2\cdot 25m}} \approx 8.9m/s$$

Answer. 8.9 m/s.