a bungee cord jumper jumps off the new river bridge. the bridge is 876 feet high. How fast is the jumper falling when he reaches 200 feet above the ground?
V^2 = Vo^2 + 2g*d
V^2 = 0 + 64*(876-200)
V^2 = 43,264
V = 208 Ft/s
To determine the speed at which the bungee jumper is falling when he reaches a height of 200 feet above the ground, we can apply the principles of physics.
First, we need to calculate the potential energy of the bungee jumper when he is at the top of the bridge, 876 feet above the ground. The potential energy is given by the formula:
Potential Energy = mass * gravitational acceleration * height
In this case, the height is 876 feet. The gravitational acceleration is approximately 32.2 feet per second squared. However, we don't have information about the mass of the bungee jumper, so we won't be able to calculate the exact potential energy.
Next, to find the speed at 200 feet above the ground, we can use the principle of conservation of mechanical energy. According to this principle, the sum of the kinetic and potential energies is constant throughout the entire fall. When the bungee jumper reaches a height of 200 feet, all the potential energy has been converted to kinetic energy.
The kinetic energy can be calculated using the following formula:
Kinetic Energy = (1/2) * mass * velocity^2
Since we don't have the mass of the bungee jumper, we won't be able to calculate the exact velocity. However, we can still determine the relative velocity by taking the square root of the ratio of the potential energy at 876 feet to the potential energy at 200 feet.
Relative Velocity = sqrt((potential energy at 876 feet / potential energy at 200 feet) * gravitational acceleration * (2 * (height at 876 feet - height at 200 feet)))
Please note that these calculations assume ideal conditions and neglect other factors like air resistance, the elasticity of the bungee cord, and the initial velocity of the jumper.