diff --git a/labs-set-2/template.pdf b/labs-set-2/template.pdf
index 0548982..aa160e8 100644
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diff --git a/labs-set-2/template.tex b/labs-set-2/template.tex
index 82ecef5..18bde53 100644
--- a/labs-set-2/template.tex
+++ b/labs-set-2/template.tex
@@ -238,4 +238,69 @@ This corresponds with answer choice B.
   \end{align*}
 }
 
+\qs{}{
+  \begin{align*}
+    \vec{v}_0=\langle 75\cos{\alpha},75\sin{\alpha} \rangle \quad \vec{a}=\langle 0,-32 \rangle 
+  \end{align*}
+  This corresponds with answer choice A.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}_0=\langle 800;34\degree \rangle \quad \vec{a}=\langle 0,-9.8 \rangle \\
+    \vec{r}\cdot\hat{i}=800\cos{34\degree}t \; \implies \frac{20000}{800\cdot\cos{34\degree}}=t_h=30.1554487126\;\mathrm{s} \\
+    \vec{r}(t_h)\cdot\hat{j}=800\sin{34\degree}30.1554487126-\tfrac{1}{2}9.8\cdot30.1554487126^2 \\
+    \vec{r}(t_h)\cdot\hat{j}=9034.35001026\ge 0
+  \end{align*}
+  $\therfore$ The answer is B.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}(t)=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=\langle 585\tfrac{\sqrt{2}}{2},585\tfrac{\sqrt{2}}{2}-9.8t \rangle \\
+    \vec{v}(t)\cdot\hat{j}=0 \; \implies t_h=\frac{585\tfrac{\sqrt{2}}{2}}{9.8} \implies t_h=42.2099456116 \\
+    \vec{r}(t_h)\cdot\hat{j}=585\tfrac{\sqrt{2}}{2}t_h-\tfrac{1}{2}9.8t_h^2=8730.22959184 \; \mathrm{m}
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    12=\|\vec{v}\|\cos{33\degree}t \quad 0=\|\vec{v}\|\sin{33\degree}t -\tfrac{1}{2}9.8t^2 \\
+    \text{Computationally solving this system of equations: } \; \|\vec{v}\| = 11.34589
+  \end{align*}
+    This corresponds with answewr choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{r}(t)\cdot\hat{j}=-\tfrac{1}{2}32t^2+115\sin{44\degree}t+6.6 \\
+    \vec{r}(4)=70.1428504111 \;\mathrm{ft}
+  \end{align*} 
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{r}(t)\cdot\hat{j}=-\tfrac{1}{2}32t^2+47\sin{39\degree}t+6.1 \\
+    \vec{r}(t_f)\cdot \hat{j}=0 \; t_f=2.03589\;\mathrm{s} \\
+    \vec{r}(t_f)\cdot\hat{i}=2.03589\cdot47\cos{39\degree}=74.3626334991\;\mathrm{ft}
+  \end{align*} 
+  This corresponds with answer choice C.
+}
+\qs{}{
+  \begin{align*}
+    0=30+39\sin{26\degree}t-\tfrac{1}{2}32t^2 \\
+    \implies t=2.00411 \;\mathrm{s}
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    x_{max}=\frac{\|\vec{v}_0\|^2\sin{2\alpha}}{g}=14000=\frac{380^2\sin{2\alpha}}{9.8} \\
+  \implies \alpha = \tfrac{1}{2}\arcsin{\frac{14000\cdot9.8}{380^2}}
+  \end{align*}
+  This corresponds with answer choice C.
+}
 \end{document}