A worker at the top of a 600-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?
(1/2) m v^2 = m g h
so
v = sqrt ( 2 g h )
V^2 = Vo^2+2g*h = 0+19.6*800 = 15680
V = 125 m/s.
To find the speed of the tool just before it hits the ground, we can use the principle of conservation of energy.
First, let's find the potential energy (PE) of the tool at the top of the tower. The potential energy is given by:
PE = m * g * h
where m is the mass of the tool, g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth), and h is the height of the tower (600 m in this case).
Next, let's find the kinetic energy (KE) of the tool just before it hits the ground. The kinetic energy is given by:
KE = (1/2) * m * v^2
where v is the speed of the tool just before it hits the ground.
According to the conservation of energy, the potential energy at the top of the tower should be equal to the kinetic energy just before hitting the ground:
PE = KE
Setting the two equations equal to each other, we have:
m * g * h = (1/2) * m * v^2
Simplifying and solving for v, we get:
v = sqrt(2 * g * h)
Plugging in the values, we have:
v = sqrt(2 * 9.8 m/s^2 * 600 m)
= sqrt(11760)
≈ 108.38 m/s
Therefore, the tool is moving at approximately 108.38 m/s just before it hits the ground.