Question

A worker at the top of a 588-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

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

$mgh=\frac{1}{2}mv^2$

where we have

$g=9.8 m/s^2\\h=588 m$

And solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s$

Learn more about kinetic energy and potential energy:

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