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To apply the law of conservation of energy to an object launched upward in Earth's gravitational field.
In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.

In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K=(1/2)mv^2 and its gravitational potential energy U=mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation

K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,

where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.

Using conservation of energy, find the maximum height h_max to which the object will rise

Answer :

Answer:

The maximum height is [tex]h_{max} = \frac{v^2}{2g}[/tex]

Explanation:

From the question we are given that

      [tex]K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,[/tex]

     and [tex]K=(1/2)mv^2[/tex]

     while [tex]U=mgh[/tex]

 Now at the minimum height the kinetic energy is maximum and the potential energy is 0

     [tex]K_i \ is \ max[/tex]

     [tex]U_i = 0[/tex]

At maximum height   [tex]h_{max}[/tex]

           The  kinetic energy is 0 and  kinetic energy is

Hence the above equation

                   [tex]K_i = U_f[/tex]

                  [tex]\frac{1}{2}mv^2 = mgh_{max}[/tex]

Making  [tex]h_{max}[/tex] the subject of the formula we have

              [tex]h_{max} = \frac{v^2}{2g}[/tex]

                 

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