it sounds like the water (above-ground pool i guess, lol) is at the same height at the platform. i could be wrong about that, but not about how to solve it:
(0.500, 6.00) is the vector. 0.5i + 6j.
her center will rise as much as any part of her body. so that would be (df - di).
her maximum height is when her upward velocity is 0. her velocity would then be downwards due to gravity. this means were going to plug vf = 0 in the formula.
the actual formula is: vf^2 = vi^2 + 2a(df - d1)
in the x direction, there is no acceleration, a_x = 0, so the formula becomes: vf_x^2 = vi_x^2
in the y direction, there is acceleration due to gravity a_y = g, so it becomes what your prof gave: vf_y^2 = vi_y^2 - 2g(df_y - di_y)
since the velocity in x and y directions are independent, we can look at them individually. we can totally ignore velocity in the x-direction for this problem. we only use the y component [6 m/s upwards]
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