Answer to Question #87697 in Mechanics | Relativity for G

Question

A worker at the top of a 526-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligible, how fast (in m/s) is the tool going just before it hits the ground?

Answer 1

We can find the final velocity of the tool just before it hits the ground from the kinematic equation:

$$v_f^2 = v_0^2 + 2gh,$$

here, $v_0 = 0$ is the initial velocity of the tool, $v_f$ is the final velocity of the tool just before it hits the ground, $g$ is the acceleration due to gravity, $h$ is the height of the television transmitting tower.

Then, we get:

$$v_f = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \dfrac{m}{s^2} \cdot 526 m} = 101.5 \dfrac{m}{s}.$$

Answer:

$v_f = 101.5 \dfrac{m}{s}.$