an object is thrown upward with an initial velocity 'u' m/s. derive an expression for the height 'h' at which its kinetic energy becomes equal to its potential energy.
h=(u²- v²)/2g
v²=u²- 2gh,
KE=m•v²/2= m(u²- 2gh)/2,
KE=PE =>
m(u²- 2gh)/2 =mgh,
(u²- 2gh)/2= gh,
u²=4gh,
h= u²/4g.
To derive an expression for the height at which the kinetic energy becomes equal to the potential energy, we can start by setting up the equation for both forms of energy.
First, let's define the variables:
- u = initial velocity of the object (in m/s)
- h = height at which the kinetic energy equals 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 the object when it is at a height h is given by:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's set up the equation and solve for h:
Equating kinetic energy and potential energy:
(1/2) * m * u^2 = m * g * h
Cancelling out the mass (m):
(1/2) * u^2 = g * h
Rearranging the equation to isolate h:
h = (1/2) * u^2 / g
So, the expression for the height at which the kinetic energy becomes equal to the potential energy is:
h = (1/2) * u^2 / g
This equation allows you to calculate the height at which the object's kinetic energy is equal to its potential energy based on its initial velocity.