1. A fish swimming in a horizontal plane has velocity overrightarrow (v)_(i)=(4hat (i)+hat (j))m/s at a point in the ocean where the position relative to a certain rock is overrightarrow (r)_(i)=(10hat (i)-4hat (j)) m. After the fish swims with constant acceleration for 20s.its velocity is overrightarrow (v)_(f)=(20hat (i)-5hat (j))m/s a) Find the acceleration of the fish b) If the fish maintains this constant acceleration, where is it at t=25s 2. A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection?

1. a) To find the acceleration of the fish, we can use the equation of motion:<br /><br />$\overrightarrow{v}_f = \overrightarrow{v}_i + \overrightarrow{a} \cdot t$<br /><br />where $\overrightarrow{v}_f$ is the final velocity, $\overrightarrow{v}_i$ is the initial velocity, $\overrightarrow{a}$ is the acceleration, and $t$ is.<br /><br />Given that $\overrightarrow{v}_i = (4\hat{i} + \hat{j}) \, \text{m/s}$, $\overrightarrow{v}_f = (20\hat{i} - 5\hat{j}) \, \text{m/s}$, and $t = 20 \, \text{s}$, we can substitute these values into the equation:<br /><br />$(20\hat{i} - 5\hat{j}) = (4\hat{i} + \hat{j}) + \overrightarrow{a} \cdot 20$<br /><br />Solving for $\overrightarrow{a}$, we get:<br /><br />$\overrightarrow{a} = \frac{(20\hat{i} - 5\hat{j}) - (4\hat{i} +{j})}{20} = \frac{16\hat{i} - 6\hat{j}}{20} = \frac{4}{5}\hat{i} - \frac{3}{10}\hat{j} \, \text{m/s}^2$<br /><br />Therefore, the acceleration of the fish is $\overrightarrow{a} = \frac{4}{5}\hat{i} - \frac{3}{10}\hat{j} \, \text{m/s}^2$.<br /><br />b) To find the position of the fish at $t = 25 \, \text{s}$, we can use the equation of motion:<br /><br />$\overrightarrow{r}_f = \overrightarrow{r}_i + \overrightarrow{v}_i \cdot t + \frac{1}{2} \overrightarrow{a} \cdot t^2$<br /><br />where $\overrightarrow{r}_f$ is the final position, $\overrightarrow{r}_i$ is the initial position, $\overrightarrow{v}_i$ is the initial velocity, $\overrightarrow{a}$ is the acceleration, and $t$ is the time.<br /><br />Given that $\overrightarrow{r}_i = (10\hat{i} - 4\hat{j}) \, \text{m}$, $\overrightarrow{v}_i = (4\hat{i} + \hat{j}) \, \text{m/s}$, $\overrightarrow{a} = \frac{4}{5}\hat{i} - \frac{3}{10}\hat{j} \, \text{m/s2$, and $t = 25 \, \text{s}$, we can substitute these values into the equation:<br /><br />$\overrightarrow{r}_f = (10\hat{i} - 4\hat{j}) + (4\hat{i} + \hat{j}) \cdot 25 + \frac{1}{2} \left(\frac{4}{5}\hat{i} - \frac{3}{10}\hat{j}\right) \cdot 25^2$<br /><br />Solving for $\overrightarrow{r}_f$, we get:<br /><br />$\overrightarrow{r}_f = (10\hat{i} - 4\hat{j}) + (100\hat{i} + 25\hat{j}) + \frac{1}{2} \left(\frac{4}{5}\hat{i} - \frac{3}{10}\hat{j}\right) \cdot 625$<br /><br />$\overrightarrow{r}_f = (110\hat{i} + 21\hat{j}) + \frac{1}{2} \left(\frac{4}{5} \cdot 625\hat{i} - \frac{3}{10} \cdot 625\hat{j}\right)$<br /><br />$\overrightarrow{r}_f = (110\hat{i} + 21\hat{j}) + 1250\hat{i} - 187.5\hat{j}$<br /><br />$\overrightarrow{r}_f = 1360\hat{i} - 166.5\hat{j} \, \text{m}$<br /><br />Therefore, the position of the fish at $t = 25 \, \text{s}$ is $\overrightarrow{r}_f = 1360\hat{i} - 166.5\hat{j} \, \text{m}$.<br /><br />2. Let the angle of projection be $\theta$. The horizontal range of the projectile is given by $R = \frac{v^2 \sin(2\theta)}{g}$, where $v$ is the initial velocity and $g$ is the acceleration due to