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Nataly [62]
3 years ago
8

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

le, how fast (in m/s) is the tool going just before it hits the ground?
Physics
1 answer:
liq [111]3 years ago
3 0

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:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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