An object is in motion according to the conditions:
$a(t)= \langle 0, 0,−32\rangle$, and $v(0)= \langle50, 0, 50\sqrt{3}\rangle$, and $r(0)=\langle 0, 0, 0\rangle$. Find the velocity and position functions for this motion. What is the maximum height (z value) that this object reaches?
Here is what I did:
$\overrightarrow v(t)= \int \overrightarrow a(t)dt = \int dt + \int dt + \int-32dt = C_1 + C_2 + -32t + C_3 = \langle C_1, C_2, -32t + C_3 \rangle$
So I assume now I solve for my constants. The first two are simple:
$\overrightarrow v(0) = \langle 50, 0, 50\sqrt{3}\rangle$
$C_1 = 50$, $C_2 = 0$, $C_3 = 50\sqrt{3}+32t$
$\overrightarrow v(t) = \langle 50, 0, 50\sqrt{3}+32t \rangle$
I'm not sure if $C_3$ is correct? Assuming it is, I then just take the integral of my new $\overrightarrow v(t)$, right?
$\overrightarrow r(t) = \int \overrightarrow v(t)dt = \int 50 dt + \int dt + \int 50\sqrt{3}dt = 50t+C_4 + C_5 + 50\sqrt{3}t + C_6 = \langle 50t+C_4, C_5, 50\sqrt{3}t+C_6\rangle $
Solving for my constants I get:
$\overrightarrow r(0) = \langle 0,0,0\rangle $
$C_4=-50t$, $C_5=0$, $C_6 = -50\sqrt{3}$
$\overrightarrow r(t) = \langle -50t, 0, -50\sqrt{3}\rangle$
Is this method correct? If it is, how do I find the maximum height the z value reaches?