2. A body fired vertically up reaches a maximum height of 120 m. The location at which the kinetic energy is 75% of the potential energy is...

For max height : mvₒ²/2 = m•g•H,

Initial velocity
vₒ=sqrt(2•g•H) = sqrt(2•9.8•120)=48.5 m/s.

mvₒ²/2 = mgh + mv²/2,
mv²/2 = mvₒ²/2 – mgh =
=m•g•H -0.75 m•g•H=0.25 m•g•H

mv²/2=0.25mgH
v =sqrt(2•0.25•g•H) = sqrt(0.5•9.8•120)=24.2 m/s.
h =(vₒ²-v²)/2•g =(48.5²-24.2²)/2•9.8=90.1 m.

To find the location at which the kinetic energy is 75% of the potential energy, we need to calculate the potential energy and kinetic energy at the maximum height.

Let's assume the mass of the body is m.
The potential energy (PE) at the maximum height is equal to the initial kinetic energy (KE0) when the body was fired vertically up.

Given that the maximum height (h) is 120 m, the potential energy at that height (PE_max) is given by:
PE_max = m * g * h

Since the kinetic energy is equal to 75% of the potential energy, we can write:
KE_max = 0.75 * PE_max
KE_max = 0.75 * m * g * h

Now, to calculate the location at which the kinetic energy is 75% of the potential energy, we need to find the height (h1) at which the kinetic energy (KE1) is equal to 0.75 * KE_max.

KE1 = 0.75 * KE_max

Now, let's assume the height at this location is h1.

At height h1, the kinetic energy is given by:
KE1 = m * g * h1

Equating the two equations for KE1, we get:
m * g * h1 = 0.75 * m * g * h

Simplifying the equation, we find:
h1 = 0.75 * h

Substituting the maximum height (h = 120 m), we get:
h1 = 0.75 * 120
h1 = 90 m

Therefore, the location at which the kinetic energy is 75% of the potential energy is at a height of 90 meters.

To find the location where the kinetic energy is 75% of the potential energy, we need to understand the relationship between kinetic energy (KE) and potential energy (PE) for an object undergoing vertical motion.

When a body is fired vertically up, its total mechanical energy is conserved throughout its trajectory. This means that the sum of kinetic energy and potential energy remains constant.

At the highest point of its trajectory (when it reaches the maximum height), the kinetic energy is zero because the body comes to a momentary rest before falling back down due to gravity. At this point, all the total mechanical energy is in the form of potential energy.

Therefore, at the maximum height, the potential energy is equal to the total mechanical energy, and the kinetic energy is zero.

Now let's consider the point where the kinetic energy becomes 75% of the potential energy. To do this, we need to find the corresponding potential energy at that point.

Since the total mechanical energy is conserved, we know that:

Potential Energy + Kinetic Energy = Total Mechanical Energy

At the maximum height, the potential energy is equal to the total mechanical energy:

Potential Energy (Max Height) = Total Mechanical Energy

Let's denote the potential energy at the maximum height as PE_max.

We can rewrite the equation as:

PE_max + 0 (since KE = 0 at the maximum height) = Total Mechanical Energy

Now, let's consider the point where the kinetic energy is 75% of the potential energy. Denote the potential energy at this point as PE_75 and the corresponding kinetic energy as KE_75. We know that:

KE_75 = 0.75 * PE_75 (75% of the potential energy)

Since the total mechanical energy is conserved, we have:

PE_75 + KE_75 = Total Mechanical Energy

Substituting the equation for KE_75, we get:

PE_75 + (0.75 * PE_75) = Total Mechanical Energy

Simplifying the equation, we find:

1.75 * PE_75 = Total Mechanical Energy

However, we do not know the value for the total mechanical energy, so we need additional information to proceed.

To find the location where the kinetic energy is 75% of the potential energy, we need to know the relationship between potential energy and height. If we are given this relationship (e.g., the potential energy as a function of height), we can proceed with solving for the location.