Answer
$−0.2099m$
Work Step by Step
Given that the speed is doubled at impact point that is the speed, we determined in part 'c' is doubled
$v_i=2\times 3.47=6.94\frac{m}{s}$
$W_{net}=\Delta{K}=0-\frac{1}{2}mv_i^2$
Also spring potential and compression kinetic energy collectively result in $W_{net}$ that is
$W_{net}=mgy+\frac{1}{2}(-ky^2)$
$-\frac{1}{2}mv^2=mgy-\frac{1}{2}ky^2$
$-\frac{1}{2}mv^2-mgy+\frac{1}{2}ky^2=0$
$\frac{1}{2}ky^2-mgy-\frac{1}{2}mv^2=0$
Applying quadratic formula to find the value of y i.e compression in spring
$y=\frac{+mg±\sqrt (mg^2)-4\times\frac{1k}{2}-(\frac{1}{2}mv_i^2)}{2\frac{1}{2}k}$
After putting the values you will get
$y=−0.2099m$
