A ball is raised to a height of 30.0m above the ground. What would its velocity be when it has fallen to a height of 15.0m above the ground. (You must solve this as a conservation of energy problem, and can do it in one step!) Any help would be appreciated..
gravitational potential energy becomes kinetic energy
m g h = 1/2 m v^2
v = √(2 g h) = √(2 * 9.81 * 15.0)
To solve this problem using the conservation of energy, we can assume that the ball only experiences gravitational potential energy and kinetic energy.
During the ball's initial position at a height of 30.0m, all of its energy is in the form of gravitational potential energy. When it falls to a height of 15.0m, some of this potential energy is converted into kinetic energy.
According to the law of conservation of energy, the initial gravitational potential energy of the ball is equal to the sum of its final gravitational potential energy and kinetic energy.
At the initial position:
Potential Energy (initial) = m * g * h (where m is the mass of the ball, g is the acceleration due to gravity, and h is the height above the ground)
At the final position:
Potential Energy (final) + Kinetic Energy (final) = m * g * h
Since the mass and acceleration due to gravity are constant, we can disregard them and rewrite the equation as:
Potential Energy (initial) = Potential Energy (final) + Kinetic Energy (final)
Substituting the given values:
m * g * 30.0m = m * g * 15.0m + Kinetic Energy (final)
Canceling out the mass and acceleration due to gravity:
30.0m = 15.0m + Kinetic Energy (final)
Rearranging the equation:
Kinetic Energy (final) = 30.0m - 15.0m = 15.0m
Since the kinetic energy is equal to one-half the mass times the velocity squared, we can solve for the final velocity:
(1/2) * m * v^2 = 15.0m
Simplifying the equation:
v^2 = 30.0
v = √(30.0)
Therefore, the velocity of the ball when it has fallen to a height of 15.0m above the ground is approximately equal to the square root of 30.0.
To solve this problem using the principle of conservation of energy, we need to consider the initial and final potential and kinetic energy of the ball.
We know that the ball is initially at a height of 30.0m above the ground. At this height, the ball has only potential energy given by the equation:
Potential energy (PE) = mass (m) * gravity (g) * height (h)
The potential energy of the ball at this height can be calculated as:
PE_initial = m * g * h_initial
Now, when the ball falls to a height of 15.0m above the ground, it still has potential energy given by:
PE_final = m * g * h_final
According to the principle of conservation of energy, the total energy of the system remains constant. This means that the initial potential energy of the ball should be equal to the final potential energy plus the final kinetic energy. Mathematically, this can be shown as:
PE_initial = PE_final + KE_final
Since the ball is falling towards the ground, its initial kinetic energy is zero. Therefore, the equation becomes:
m * g * h_initial = m * g * h_final + KE_final
To find the velocity (v) when the ball has fallen to a height of 15.0m, we need to relate the final kinetic energy (KE_final) to the velocity using the equation:
KE_final = 0.5 * m * v^2
Substituting this into the previous equation, we get:
m * g * h_initial = m * g * h_final + 0.5 * m * v^2
We can cancel out the mass (m) on both sides of the equation:
g * h_initial = g * h_final + 0.5 * v^2
Now, let's rearrange the equation to solve for the velocity (v):
0.5 * v^2 = g * (h_initial - h_final)
Dividing both sides of the equation by 0.5 gives:
v^2 = 2 * g * (h_initial - h_final)
Finally, taking the square root of both sides of the equation, we can solve for the velocity (v):
v = √(2 * g * (h_initial - h_final))
Substituting the given values:
v = √(2 * 9.8 m/s^2 * (30.0m - 15.0m))
Calculating this expression, we get:
v ≈ √(2 * 9.8 * 15.0)
v ≈ √(294)
v ≈ 17.14 m/s
Therefore, the velocity of the ball when it has fallen to a height of 15.0m above the ground is approximately 17.14 m/s.