Get Best Essay Written by US Essay Writers  ## Question:

What Is The Speed u Of The Object At The Height Of 1-2h Max

## Introduction:

When an object is thrown upward, it gains potential energy, which is converted to kinetic energy as it falls back towards the ground. The maximum height reached by the object is determined by its initial velocity and the acceleration due to gravity. The speed of the object at any given height can be calculated using the laws of motion and energy conservation principles.

## Calculation Of Speed At The Height Of 1/2 hmax:

To calculate the speed of the object at the height of 1/2 hmax, we need to first determine the maximum height reached by the object. The maximum height can be calculated using the equation:

hmax = (v0^2)/(2g)

Where v0 is the initial velocity and g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on the Earth's surface.

Once we have determined the maximum height, we can calculate the speed of the object at the height of 1/2 hmax using the conservation of energy principle. At any given height, the total energy of the object is the sum of its kinetic and potential energies. The equation for total energy is:

E = K + U

Where E is the total energy, K is the kinetic energy, and U is the potential energy.

At the height of 1/2 hmax, the potential energy of the object is half of its maximum potential energy, which can be calculated using the equation:

Umax = mghmax

Where m is the mass of the object, g is the acceleration due to gravity, and hmax is the maximum height.

Therefore, the potential energy at the height of 1/2 hmax is:

U = 1/2 mghmax

The kinetic energy at the height of 1/2 hmax can be calculated using the conservation of energy principle, which states that the total energy of the object is constant throughout its motion. Therefore, we can write:

E = constant

K + U = constant

At the height of 1/2 hmax, the potential energy is 1/2 of its maximum value and the kinetic energy is unknown. We can express this as:

K + 1/2 mghmax = constant

Solving for K, we get:

K = constant - 1/2 mghmax

Substituting the value of U and simplifying, we get:

K = 1/2 mghmax - 1/2 mgh

The speed of the object at the height of 1/2 hmax can be calculated using the equation for kinetic energy:

K = 1/2 mv^2

Substituting the value of K and simplifying, we get:

1/2 mv^2 = 1/2 mghmax - 1/2 mgh

v^2 = 2g(hmax - h)

v = sqrt(2g(hmax - h))

## Conclusion:

In conclusion, the speed of the object at the height of 1/2 hmax can be calculated using the conservation of energy principle and the equations for potential and kinetic energy. The maximum height reached by the object can be calculated using the equation for maximum height, which is determined by the initial velocity and acceleration due to gravity. The speed at any given height can be calculated using the above equations and principles. Hurry and fill the order form