If an object is dropped from 10 m above the ground, what is the height at which its kinetic energy and potential energy are equa
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Hi there!
We can use the following kinematic equation:
vf = final velocity (? m/s)
vi = intial velocity (0 m/s)
a = acceleration (5 m/s²)
d = displacement (8 m)
Plug in the givens and solve.
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity:
where
is the final velocity
is the initial velocity
is the acceleration due to gravity 
is the time
2. Kinematic equation for distance:
where
is the distance
is the initial velocity
is the final velocity
is the acceleration due to gravity
is the time
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
. We also know that it initial speed is 31.21 ft/s, so 
. Lets replace those values in our formula to find :
seconds
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
ft
Now we can add the height of the building and the maximum height of the object:
ft
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
or 
Since time cannot be negative,
is the time it takes the object to fall to the ground.
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
ft/s
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s
1. Kinematic equation for final velocity:
where
2. Kinematic equation for distance:
where
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
Now we can add the height of the building and the maximum height of the object:
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
Since time cannot be negative,
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.
The energy of the emitted photon is given by:
where
h is the Planck constant
f is the photon frequency
Substituting
, we find
This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
The energy of the emitted photon is given by:
where
h is the Planck constant
f is the photon frequency
Substituting
This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)