Answer:
This question can be answered by using conversation of energy.
[tex]K_1 + U_1 = K_2 + U_2[/tex]
[tex]\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2[/tex]
[tex]\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s[/tex]
Explanation:
Note that we take [tex]K_2 = 0[/tex] because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.