1. **State the problem:** A ball is released with an initial speed of $6\ \text{m/s}$ at an angle of $45^\circ$ from a height of $2.5\ \text{m}$. We want to determine if the ball will pass through a loop located $7\ \text{m}$ horizontally away and $3.05\ \text{m}$ high. 2. **Relevant formulas:** - Horizontal position: $$x = v_0 \cos(\theta) t$$ - Vertical position: $$y = y_0 + v_0 \sin(\theta) t - \frac{1}{2} g t^2$$ where $v_0 = 6$, $\theta = 45^\circ$, $y_0 = 2.5$, and $g = 9.8\ \text{m/s}^2$. 3. **Find time $t$ when ball reaches horizontal distance $7$ m:** $$7 = 6 \cos(45^\circ) t$$ $$7 = 6 \times \frac{\sqrt{2}}{2} t$$ $$7 = 3\sqrt{2} t$$ $$t = \frac{7}{3\sqrt{2}}$$ 4. **Calculate vertical position $y$ at time $t$:** $$y = 2.5 + 6 \sin(45^\circ) \times \frac{7}{3\sqrt{2}} - \frac{1}{2} \times 9.8 \times \left(\frac{7}{3\sqrt{2}}\right)^2$$ Since $\sin(45^\circ) = \frac{\sqrt{2}}{2}$: $$y = 2.5 + 6 \times \frac{\sqrt{2}}{2} \times \frac{7}{3\sqrt{2}} - 4.9 \times \frac{49}{18}$$ Simplify numerator and denominator: $$y = 2.5 + 3 \times \frac{7}{3} - 4.9 \times \frac{49}{18}$$ $$y = 2.5 + 7 - 4.9 \times 2.7222$$ $$y = 9.5 - 13.333$$ $$y = -3.833$$ 5. **Interpretation:** The vertical position at $7\ \text{m}$ is approximately $-3.83\ \text{m}$, which is below ground level and certainly below the loop height of $3.05\ \text{m}$. Therefore, the ball will not go through the loop. **Final answer:** The ball does not reach the height of the loop at $7\ \text{m}$ horizontally; it falls short and will not go through the loop.