An object is dropped from a height of 12m.At what height is the rock kinetic energy and its potential energy be equal?

assume potential energy = 0 when h = 0

U = m g h
U top = total energy = m (9.8 * 12)
=

U + Ke = U top
so
if U = Ke then
U = U top /2 = m (9.8 * 6)
so
at h = 6 meters

In other words, when it loses half its original potential energy, the Ke = U and that is halfway down.

Well, well, well, let's calculate the height at which the rock's kinetic energy and potential energy become equal!

First, let's consider the initial potential energy when the object is dropped from a height of 12m. We can use the equation for potential energy: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Now, once the rock reaches its maximum kinetic energy, all of the potential energy would be converted to kinetic energy. So, we'll set the potential energy equal to the kinetic energy and solve for the height.

Let's assume the mass of the rock is 1kg for simplicity. The acceleration due to gravity is approximately 9.8 m/s^2. Using the equation PE = KE, we get mgh = (1/2)mv^2, where v is the velocity at the maximum kinetic energy.

Simplify and cancel out the mass and velocity: gh = (1/2)v^2

Since we want to find the height, let's rearrange the equation: h = (1/2)v^2 / g

Now, we need to find the velocity at the maximum kinetic energy. When the object is dropped, it accelerates until it reaches the ground, so we can use the equation v^2 = u^2 + 2as, where u is the initial velocity, s is the displacement, and a is the acceleration. Since the object is dropped, the initial velocity is zero, and the displacement is the height h.

Substituting the known values, v^2 = 0^2 + 2 * 9.8 * 12

Simplify and solve for v^2: v^2 = 2 * 9.8 * 12

Finally, plug this value of v^2 back into our height equation: h = (1/2) * (2 * 9.8 * 12) / 9.8

Cancelling out the 9.8, h = (1/2) * 12

Drumroll, please... h = 6m!

Voila! At a height of 6 meters, the rock's kinetic energy and potential energy will be equal. Just don't forget, when you reach that height, you might bump into some funny clowns!

To determine at what height the kinetic energy and potential energy of an object are equal, we need to equate the formulas for each type of energy.

The potential energy (PE) of an object is given by the formula:

PE = m * g * h

where:
m = mass of the object
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2) * m * v^2

where:
m = mass of the object
v = velocity

In this case, the mass of the object is not given and is not required to find the height.

Let's set the potential energy equal to the kinetic energy:

PE = KE

m * g * h = (1/2) * m * v^2

Cancel out the mass on both sides:

g * h = (1/2) * v^2

Now, let's substitute the values for acceleration due to gravity and rearrange the equation:

9.8 * h = (1/2) * v^2

Now, let's plug in the known values. Since the object is dropped, its initial velocity is 0:

9.8 * h = (1/2) * 0^2

This simplifies to:

9.8 * h = 0

Therefore, the height at which the rock's kinetic energy and potential energy are equal is 0 meters.

To determine the height at which the kinetic energy and potential energy of the object are equal, we need to equate the formulas for kinetic energy and potential energy.

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2) * m * v^2

where m is the mass of the object and v is its velocity.

The potential energy (PE) of an object near the Earth's surface is given by the formula:

PE = m * g * h

where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the reference point.

At the height where kinetic energy and potential energy are equal, KE = PE. Therefore, we can set up the equation:

(1/2) * m * v^2 = m * g * h

To simplify the equation, we can cancel out the mass term on both sides:

(1/2) * v^2 = g * h

Next, we can isolate h by dividing both sides of the equation by g:

h = (1/2) * (v^2 / g)

Now, we can substitute the given values into the equation to find the height:

v = 0 (since the object is dropped, its initial velocity is zero)
g = 9.8 m/s^2
h = ?

Plugging in these values into the equation:

h = (1/2) * (0 / 9.8)
h = 0

So, at a height of 0 meters (or when it reaches the ground), the rock's kinetic energy and potential energy are equal.