We know that if a particle moves along a curve in a plane, then the expression for the velocity of the particle is given by,

\( \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta} \)

[ To know the derivation for the velocity (CLICK HERE) ]

and the expression for the acceleration is given by,

\( \vec{a}=(\ddot{r}-r\ {\dot{\theta}}^2)\hat{r}+({r}\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta} \)

[ To know the derivation for the acceleration (CLICK HERE) ]

Given, 

\( \displaystyle{r=3t-\frac{1}{30}t^2} \)

\( \displaystyle{\dot{r}=\frac{dr}{dt}=3-\frac{1}{30}\times{2}t} \)

\( or,\ \displaystyle{\dot{r}=3-\frac{1}{15}t} \)

Again,

\( \displaystyle{\ddot{r}=\frac{d\dot{r}}{dt}=-\frac{1}{15}} \)

Now at \( t=30\ sec \),

\( \displaystyle{r=3\times{30}-\frac{1}{30}\times{30}^2} \)

\( or,\ r=90-30=60\ m \)

\( \displaystyle{\dot{r}=3-\frac{1}{15}\times{30}}=1\ m/sec \)

and, \( \displaystyle{\ddot{r}=\frac{d\dot{r}}{dt}=-\frac{1}{15}}\ m/{sec}^2 \)

Also given,

\( {\theta}^2=1600-t^2 \)

\( or,\ \theta=\sqrt{1600-t^2} \)

\( \displaystyle{\dot{\theta}=\frac{d\theta}{dt}=\frac{d}{dt}[\sqrt{1600-t^2}]} \)

\( or,\ \displaystyle{\dot{\theta}=\frac{(-2t)}{2[\sqrt{1600-t^2}]}} \)

\( or,\ \displaystyle{\dot{\theta}=-\frac{(t)}{[\sqrt{1600-t^2}]}} \)

Again,

\( \displaystyle{\ddot{\theta}=\frac{d\dot{\theta}}{dt}=\frac{d}{dt}[-\frac{(t)}{[\sqrt{1600-t^2}]}]} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{\sqrt{1600-t^2}-t\frac{-2t}{2\sqrt{1600-t^2}}}{(1600-t^2)}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{\sqrt{1600-t^2}+ t\frac{t}{\sqrt{1600-t^2}}}{(1600-t^2)}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{\sqrt{1600-t^2}+ \frac{t^2}{\sqrt{1600-t^2}}}{(1600-t^2)}} \)

Now at \( t=30\ sec \),

\( \theta=\sqrt{1600-30^2}=\sqrt{1600-900}\ rad \)

\( or,\ \theta=\sqrt{700}=10\sqrt{7}\ rad \)

\( \displaystyle{\dot{\theta}=-\frac{(30)}{[\sqrt{1600-30^2}]}}\ rad/sec \)

\( or,\ \displaystyle{\dot{\theta}=-\frac{30}{10\sqrt{7}}}\ rad/sec \)

\( or,\ \displaystyle{\dot{\theta}=-\frac{3}{\sqrt{7}}}\ rad/sec \)

and,

\( \displaystyle{\ddot{\theta}=-\frac{\sqrt{1600-30^2}+ \frac{30^2}{\sqrt{1600-30^2}}}{(1600-30^2)}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{10\sqrt{7}+ \frac{900}{10\sqrt{7}}}{(700)}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{700+900}{10\sqrt{7}\times700}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{1600}{7000\sqrt{7}}} \)

\( or,\ \displaystyle{\ddot{\theta}=-\frac{16}{70\sqrt{7}}} \)

So the velocity of the projectile at \( t=30\ sec \) is

\( \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta} \)

\( or,\ \vec{v}=1\hat{r}+60\times(-)\frac{3}{\sqrt{7}}\hat{\theta} \)

\( or,\ \displaystyle{\vec{v}=\hat{r}-\frac{180}{\sqrt{7}}\hat{\theta}} \)

and the acceleration of the projectile at \( t=30\ sec \),

\( \vec{a}=(\ddot{r}-r\ {\dot{\theta}}^2)\hat{r}+({r}\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta} \)

\( or,\ \displaystyle{\vec{a}=\left(-\frac{1}{15}-60\times{(-\frac{3}{\sqrt{7}})}^2\right)\hat{r}+\left(60\times\frac{-16}{70\sqrt{7}}+2\times{1}\times\frac{-3}{\sqrt{7}}\right)\hat{\theta}} \)

\( or,\ \displaystyle{\vec{a}=\left(-\frac{1}{15}-\frac{540}{7}\right)\hat{r}+\left(-\frac{96}{7\sqrt{7}}-\frac{6}{\sqrt{7}}\right)\hat{\theta}} \)

\( or,\ \displaystyle{\vec{a}=-\left(\frac{8107}{105}\right)\hat{r}-\left(\frac{138}{7\sqrt{7}}\right)\hat{\theta}} \)

\( or,\ \displaystyle{\vec{a}=-\left[\left(\frac{8107}{105}\right)\hat{r}+\left(\frac{138}{7\sqrt{7}}\right)\hat{\theta}\right]} \)