Calculus - Find the position and velocity of an object when it hits the ground - using vector

Question

If an object is thrown in the air with an initial position of $(0,0,4)$ in feet and initial velocity $(200,0,256)$ in feet per second, and the acceleration is $(0,8,-32)$. Find the position and velocity of an object when it hits the ground.

My attempt

Tangential Component:

$ \frac{\langle(200,0,256),\,(0,8,-32)\rangle}{\|(200,0,256)\|}$ = $\frac{-8192}{8\sqrt{1649}}$

Normal Component:

$ \frac{\|(2048,6400,1600)\|}{\|(200,0,256)\|}$ = $\frac{64\sqrt{11649}}{8\sqrt{1649}}$

Answer 1

Your tangential has $\vec {v_0}\cdot \vec a$ which doesn't make sense and I don't know where $\langle 2048,6400,1600 \rangle$ comes from in the normal. I would just use $\vec s=\vec{s_0}+\vec {v_0}t+\frac 12\vec a t^2$ You are given the constants on the right. Solve for $t$ that makes the height (presumably the third component) zero, then plug that in to get the first two.