Answer
$d=(5sin42^\circ)sin(\frac{\pi}{3}t)\approx3.346sin(\frac{\pi}{3}t)$
Work Step by Step
1. Model the motion with simple harmonic $d=A\ sin(kt)$
2. Based on the given conditions, we have $p=\frac{2\pi}{k}=6$, thus $k=\frac{\pi}{3}$. And the amplitude $A=5sin42^\circ\approx3.346\ ft$
3. Thus we have $d=(5sin42^\circ)sin(\frac{\pi}{3}t)\approx3.346sin(\frac{\pi}{3}t)$
