Whoever wrote that is very wrong. You need to considerthe total energy she has. At the top she has both KE and PE. Her KE is \(\displaystyle{\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}\) or \(\displaystyle\frac{{\frac{{{1}}}{{{2}}}}}{{{35}}}{\left({5}\right)}^{{2}}\) Plus her PE given amgh or \(\displaystyle{35}\times{9.8}\times{11}\). She loses energy on the way down due tofriction in the amount of \(\displaystyle{F}\times{D}\) or \(\displaystyle{15}\times{100}={1500}\). So to find V final. ADD the KE top plus the PE top and subtract 1500 and that equals \(\displaystyle{\frac{{{1}}}{{{2}}}}{m}{v}^{{2}}\) (KE only at bottom) You should get 12.4 m/s