Question

19.) A ball is thrown horizontally from the top of a building 130 m high. The ball strikes the ground 59 m horizontally from the point of release. What is the speed of the ball just before it strikes the ground ?

Answer 1

The time of falling can be found from the kinematic equation:

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

where $h = 130m$ if the height of the builing, and $g = 9.81m/s^2$ if the gravitational acceleration.

Thus, obtain:

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

The speed in veritical direction by the time $t$ is (motion with uniform acceleration):

$$v_y = v_{y0} + gt = gt = \sqrt{2gh}$$

where $v_{y0} = 0m/s$ is the initial vertical speed.

The speed in horizontal direction by the time $t$ is (motion with constant speed):

$$v_x = d/t = d\sqrt{\dfrac{g}{2h}}$$

where $d = 59m$ is the range.

The total speed is then:

$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{2gh + \dfrac{d^2g}{2h}} = \sqrt{9.81\cdot \left(2\cdot 130 + \dfrac{59^2}{2\cdot 130}\right)} \approx 51.8m/s$$

Answer. 51.8 m/s.