Question

A ball is thrown downward from the top of a 46.0 m tower with an initial speed of 10.0 m/s. Assuming negligible air resistance, what is the speed of the ball just before hitting the ground?

Answer 1

The kinematic law of the motion with the constant acceleration is:

$$d = v_0t + \dfrac{gt^2}{2}$$

where $d = 46m$ is the height of the tower, $v_0 = 10m/s$ is the initial speed of the ball and $g = 9.81 m/s^2$ is the gravitational acceleration.

Solving the quadratic equation

$$\dfrac{gt^2}{2} +v_0t - d = 0$$

for $t$, obtain the total time of falling:

$$t_{fall} = \dfrac{-v_0 + \sqrt{v_0^2 + 2gd}}{2}$$

The speed after $t_{fall}$ seconds will be:

$$v = v_0 + gt_{fall} = v_0 + \dfrac{g(\sqrt{v_0^2 + 2gd}-v_0)}{2} =\\ =10+ \dfrac{9.81(\sqrt{10^2 + 2\cdot 9.81\cdot 46}-10)}{2} \approx 116.26 m/s$$

Answer. 116.26 m/s.