A hammer falls from a scaffold on a building 50.0 m above the ground. Find its speed as it hits the
ground.
V^2 = Vo^2 + 2g*d = 0 + 19.6*50 = 980
V = 31.3 m/s.
To find the speed of the hammer as it hits the ground, we can use the principles of motion and consider the conservation of energy.
First, we need to determine the potential energy the hammer has at the top of the scaffold and the kinetic energy it has just before it hits the ground.
At the top of the scaffold, the hammer only has potential energy, given by the equation:
Potential Energy = mass * gravity * height
where:
mass = mass of the hammer
gravity = acceleration due to gravity (approximately 9.8 m/s^2)
height = height of the scaffold (50.0 m in this case)
Next, we need to consider the conservation of energy. According to the principle of energy conservation, the total mechanical energy of the hammer is constant throughout its motion. At the top of the scaffold, all the energy is potential energy, and just before hitting the ground, all the energy is kinetic energy.
Therefore, the potential energy at the top of the scaffold is equal to the kinetic energy just before hitting the ground:
Potential Energy = Kinetic Energy
From the equation for kinetic energy:
Kinetic Energy = (1/2) * mass * velocity^2
where:
velocity = speed of the hammer just before hitting the ground (what we want to find)
Setting these two equations equal to each other and solving for the velocity:
mass * gravity * height = (1/2) * mass * velocity^2
canceling out the mass term:
gravity * height = (1/2) * velocity^2
isolating the velocity:
velocity^2 = 2 * gravity * height
taking the square root of both sides:
velocity = sqrt(2 * gravity * height)
Plugging in the values for gravity (9.8 m/s^2) and height (50.0 m):
velocity = sqrt(2 * 9.8 * 50.0)
velocity = sqrt(980)
velocity ≈ 31.3 m/s
Therefore, the speed of the hammer as it hits the ground is approximately 31.3 m/s.