1. **Problem Statement:** (a)(i) Express velocity $V$ as the subject of the formula $E = \frac{mv^2}{2} + mgh$. (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 $x$ given $5(x-3) = 3(x-10)$. --- ### (a)(i) Express $V$ as the subject 2. Start with the formula: $$E = \frac{mv^2}{2} + mgh$$ 3. Subtract $mgh$ from both sides: $$E - mgh = \frac{mv^2}{2}$$ 4. Multiply both sides by 2: $$2(E - mgh) = mv^2$$ 5. Divide both sides by $m$: $$\frac{2(E - mgh)}{m} = v^2$$ 6. Take the square root of both sides: $$V = \sqrt{\frac{2(E - mgh)}{m}}$$ --- ### (a)(ii) Evaluate $V$ 7. Substitute values: $$V = \sqrt{\frac{2(4900 - 20 \times 9.8 \times 15)}{20}}$$ 8. Calculate inside the parentheses: $$20 \times 9.8 \times 15 = 2940$$ 9. So: $$V = \sqrt{\frac{2(4900 - 2940)}{20}} = \sqrt{\frac{2(1960)}{20}}$$ 10. Simplify numerator: $$2 \times 1960 = 3920$$ 11. Divide: $$\frac{3920}{20} = 196$$ 12. Take square root: $$V = \sqrt{196} = 14$$ --- ### (b)(i) Find present ages 13. Let student's age be $x$, instructor's age be $3x$. 14. Fifteen years ago: $$3x - 15 = 18(x - 15)$$ 15. Expand right side: $$3x - 15 = 18x - 270$$ 16. Rearrange: $$3x - 15 - 18x + 270 = 0 \Rightarrow -15x + 255 = 0$$ 17. Solve for $x$: $$-15x = -255 \Rightarrow x = 17$$ 18. Instructor's age: $$3x = 51$$ --- ### (b)(ii) Solve for $x$ 19. Given: $$5(x - 3) = 3(x - 10)$$ 20. Expand: $$5x - 15 = 3x - 30$$ 21. Rearrange: $$5x - 3x = -30 + 15 \Rightarrow 2x = -15$$ 22. Solve: $$x = -\frac{15}{2} = -7.5$$