A boulder with a mass of 2,500 kg on a ledge 200 m above the ground falls. If the boulder's mechanical energy is conserved, what is the speed of the boulder just before it hits the ground?
1/2 m vf^2=m*g*height
solve for vf
notice mass divides out.
vf=sqrt(2gh)
How do you figure miles per hour?
what do you mean, Adam?
I meant meters per second
To find the speed of the boulder just before it hits the ground, we need to use the principle of conservation of mechanical energy. The mechanical energy of an object is the sum of its kinetic energy and potential energy.
Initially, when the boulder is on the ledge, it only has potential energy due to its height above the ground. The potential energy (PE) of an object near the surface of the Earth is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
PE = mgh
Given:
m (mass of the boulder) = 2,500 kg
g (acceleration due to gravity) = 9.8 m/s²
h (height above the ground) = 200 m
Substituting these values into the equation, we can find the initial potential energy of the boulder:
PE = (2,500 kg)(9.8 m/s²)(200 m)
= 4,900,000 kg·m²/s²
Since mechanical energy is conserved, the initial potential energy of the boulder will be equal to its final kinetic energy (KE) just before it hits the ground.
KE = 4,900,000 kg·m²/s²
The kinetic energy of an object is given by the equation KE = (1/2)mv², where m is the mass of the object and v is its velocity.
Substituting the values into the equation, we can find the final kinetic energy of the boulder just before it hits the ground:
4,900,000 kg·m²/s² = (1/2)(2,500 kg)v²
To find the velocity (v), rearrange the equation:
v² = (2 × 4,900,000 kg·m²/s²) / (2,500 kg)
= 9,800 m²/s²
Taking the square root of both sides, we can find the velocity (v):
v ≈ √(9,800 m²/s²)
≈ 99 m/s
Therefore, the speed of the boulder just before it hits the ground is approximately 99 m/s.