Answer:
[tex]K_p=6174J[/tex]
Explanation:
Given:
[tex]m=70kg[/tex],[tex]m=9m[/tex],[tex]K_E=70kg[/tex]
[tex]K_E=\frac{1}{2}*m*v^2[/tex]
[tex]1200J=\frac{1}{2}*70kg*v^2[/tex]
[tex]v=5.8m/s[/tex]
To determine the kinetic energy at the bottom is:
[tex]K_p=m*g*h[/tex]
[tex]K_p=70kg*9.8m/s^2*9m[/tex]
[tex]K_p=6174J[/tex]