A body projected vertically with a velocity “u” from the ground . It’s velocity at ¼ th of maximum height is

When a body is projected vertically upward with an initial velocity "u," it experiences a change in velocity due to the acceleration caused by gravity. To determine the velocity at one-fourth of its maximum height, we first need to understand a few key concepts related to projectile motion.

Understanding Maximum Height

The maximum height (H) reached by the body can be calculated using the formula:

• H = (u²) / (2g)

Here, "g" represents the acceleration due to gravity, approximately 9.81 m/s². This formula arises from the principle that the kinetic energy at the point of projection is converted into potential energy at the maximum height.

Calculating One-Fourth of Maximum Height

To find the height at one-fourth of the maximum height, we can use:

• Height at ¼ H = H/4 = (u²) / (8g)

Applying Kinematic Equations

Next, we can use the kinematic equations to find the velocity at this height. The relevant equation is:

• v² = u² - 2g * h

In this case, "h" is the height at one-fourth of the maximum height, which we calculated as (u²) / (8g). Substituting this value into the equation gives:

Substituting Values

Now, substituting for "h" in the kinematic equation:

• v² = u² - 2g * (u² / (8g))

This simplifies to:

• v² = u² - (u² / 4)

Combining the terms results in:

• v² = (3u²) / 4

Finding the Velocity

To find the velocity "v," we take the square root of both sides:

• v = √((3u²) / 4)

This further simplifies to:

• v = (√3 / 2) * u

Final Result

Therefore, the velocity of the body at one-fourth of its maximum height is:

• v = (√3 / 2) * u

This means that at this point in its trajectory, the body retains a significant portion of its initial velocity, specifically about 86.6% of it, as it continues to ascend before reaching its peak.