Question

To apply the law of conservation of energy to an object launched upward in Earth's gravitational field.

Answer 1

Answer:

The maximum height is $h_{max} = \frac{v^2}{2g}$

Explanation:

From the question we are given that

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

     and $K=(1/2)mv^2$

     while $U=mgh$

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

     $K_i \ is \ max$

     $U_i = 0$

At maximum height   $h_{max}$

           The  kinetic energy is 0 and  kinetic energy is

Hence the above equation

                   $K_i = U_f$

                  $\frac{1}{2}mv^2 = mgh_{max}$

Making  $h_{max}$ the subject of the formula we have

              $h_{max} = \frac{v^2}{2g}$