Answer
a) $v=28.2\frac{m}{s}$
b) $s=116.6m$
Work Step by Step
a) $\Delta h=45.0m-4.4m=40.6m$
$W=\Delta E_P=\Delta E_K$
$(55kg)(9.8\frac{m}{s^2})(40.6m)=2.19\times10^5J$
$\frac{(55kg)v^2}{2}=2.19\times10^5J$
$v=28.2\frac{m}{s}$
b) In the time it took her to go down 4.4 m, she went forward
$4.4m=\frac{(9.8\frac{m}{s^2})t^2}{2}$
$t=0.948s$
$d=(28.2\frac{m}{s})(0.948s)=26.7m$
$y_v=xtan(30^o)=15.4m$
$y_v=\frac{gx^2}{2v^2}+y_i=0.577x$
Rearranging the equation to solve for x, we have
$x^2-\frac{2(0.577)(28.2\frac{m}{s})^2}{9.8\frac{m}{s^2}}x-\frac{2(4.4m)(28.2\frac{m}{s})^2}{9.8\frac{m}{s^2}}=0$
$x^2-93.7x-715=0$
$x=101m,x=-7.09m$
$x$ is positive so the first value is our answer
From $x=s\times\cos(30^o)$, $s=\frac{x}{\cos(30^o)}=116.6m$
