Question

A rock is thrown horizontally from the top of a building with an initial speed of v = 10.1 m/s. If it lands d = 57.1 m from the base of the building, how high is the building?

Answer 1

The speed in horizontal direction remains unchanged during the falling. Thus, the time of falling is given as follows:

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

where $d = 57.1m$ is the range, and $v_h = 10.1m/s$ is the horizontal speed.

The distance travelled in vertical direction (height of the building) is given by the following kinematic equation:

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

where $g = 9.81m/s^2$ is the gravitational acceleration. Substituting the expression for $t$, obtain:

$$h = \dfrac{gd^2}{2v_h^2} = \dfrac{9.81\cdot 57.1^2}{2\cdot 10.1^2} \approx 157m$$

Answer. 157 m.