1. **Problem Statement:** (a)(i) Given the energy formula $$E = \frac{mv^2}{2} + mgh$$, express velocity $V$ as the subject. (a)(ii) Evaluate $V$ when $m=20$, $h=15$, $E=4900$, and $g=9.8$. (b)(i) An instructor is three times as old as his student. Fifteen years ago, the instructor was eighteen times as old as the student. Find their present ages. (b)(ii) Solve for $x$ in $5(x-3) = 3(x-10)$. 2. **Formulas and Rules:** - Energy formula: $$E = \frac{mv^2}{2} + mgh$$ - Algebraic manipulation to isolate $V$. - Age problem uses simultaneous equations. - Linear equation solving for $x$. 3. **Step-by-step Solutions:** **(a)(i) Express $V$ as the subject:** $$E = \frac{mv^2}{2} + mgh$$ Subtract $mgh$ from both sides: $$E - mgh = \frac{mv^2}{2}$$ Multiply both sides by 2: $$2(E - mgh) = mv^2$$ Divide both sides by $m$: $$\frac{2(E - mgh)}{m} = v^2$$ Take the square root: $$V = \sqrt{\frac{2(E - mgh)}{m}}$$ **(a)(ii) Evaluate $V$:** Substitute $m=20$, $h=15$, $E=4900$, $g=9.8$: $$V = \sqrt{\frac{2(4900 - 20 \times 9.8 \times 15)}{20}}$$ Calculate inside the parentheses: $$20 \times 9.8 \times 15 = 2940$$ So: $$V = \sqrt{\frac{2(4900 - 2940)}{20}} = \sqrt{\frac{2(1960)}{20}} = \sqrt{\frac{3920}{20}} = \sqrt{196}$$ Therefore: $$V = 14 \text{ m/s}$$ **(b)(i) Age problem:** Let student age = $x$, instructor age = $3x$. Fifteen years ago: Student age = $x - 15$, instructor age = $3x - 15$. Given: $$3x - 15 = 18(x - 15)$$ Expand right side: $$3x - 15 = 18x - 270$$ Bring terms to one side: $$3x - 15 - 18x + 270 = 0$$ $$-15x + 255 = 0$$ $$-15x = -255$$ $$x = 17$$ Student age = 17 years. Instructor age = $3 \times 17 = 51$ years. **(b)(ii) Solve for $x$:** $$5(x - 3) = 3(x - 10)$$ Expand: $$5x - 15 = 3x - 30$$ Bring terms to one side: $$5x - 15 - 3x + 30 = 0$$ $$2x + 15 = 0$$ $$2x = -15$$ $$x = -\frac{15}{2} = -7.5$$ **Final answers:** (a)(i) $$V = \sqrt{\frac{2(E - mgh)}{m}}$$ (a)(ii) $$V = 14 \text{ m/s}$$ (b)(i) Student age = 17 years, Instructor age = 51 years. (b)(ii) $$x = -7.5$$