1. **Problem Statement:** We have the height of a ball given by the equation $$h = -4 (t - 3)^2 + 46.5$$ where $h$ is the height in metres and $t$ is the time in seconds after the ball is thrown. 2. **Formula and Important Rules:** This is a quadratic function in vertex form: $$h = a(t - h)^2 + k$$ where $(h, k)$ is the vertex. - The vertex gives the maximum or minimum point of the parabola. - Since $a = -4 < 0$, the parabola opens downward, so the vertex is the maximum point. - The $y$-intercept is found by setting $t=0$. - The $x$-intercepts (roots) are found by setting $h=0$. 3. **a) Time to reach greatest height:** The vertex is at $t=3$ seconds, so the ball reaches its greatest height at $$\boxed{3 \text{ seconds}}$$. 4. **b) Greatest height reached:** The height at the vertex is $46.5$ metres, so the greatest height is $$\boxed{46.5 \text{ metres}}$$. 5. **c) Initial height (height at $t=0$):** Calculate $h$ when $t=0$: $$h = -4(0 - 3)^2 + 46.5 = -4(9) + 46.5 = -36 + 46.5 = 10.5$$ So, the ball is thrown from $$\boxed{10.5 \text{ metres}}$$. 6. **d) Height at $t=1.5$ seconds:** Calculate $h$ when $t=1.5$: $$h = -4(1.5 - 3)^2 + 46.5 = -4(-1.5)^2 + 46.5 = -4(2.25) + 46.5 = -9 + 46.5 = 37.5$$ The height at 1.5 seconds is $$\boxed{37.5 \text{ metres}}$$. 7. **e) Time when ball hits the ground ($h=0$):** Set $h=0$: $$0 = -4(t - 3)^2 + 46.5$$ Rearranged: $$4(t - 3)^2 = 46.5$$ $$ (t - 3)^2 = \frac{46.5}{4} = 11.625$$ Take square root: $$t - 3 = \pm \sqrt{11.625} \approx \pm 3.41$$ So, $$t = 3 \pm 3.41$$ Two solutions: - $$t = 3 - 3.41 = -0.41$$ (not physically meaningful since time cannot be negative) - $$t = 3 + 3.41 = 6.41$$ seconds Therefore, the ball hits the ground at $$\boxed{6.41 \text{ seconds}}$$. 8. **f) Times when ball is at 16 metres:** Set $h=16$: $$16 = -4(t - 3)^2 + 46.5$$ Rearranged: $$-4(t - 3)^2 = 16 - 46.5 = -30.5$$ $$ (t - 3)^2 = \frac{30.5}{4} = 7.625$$ Take square root: $$t - 3 = \pm \sqrt{7.625} \approx \pm 2.76$$ So, $$t = 3 \pm 2.76$$ Two times: - $$t = 3 - 2.76 = 0.24$$ seconds - $$t = 3 + 2.76 = 5.76$$ seconds Rounded to 2 decimal places, the ball is at 16 metres at $$\boxed{0.24 \text{ seconds}}$$ and $$\boxed{5.76 \text{ seconds}}$$. 9. **g) Sketch description:** The parabola opens downward with vertex at $(3, 46.5)$. - The ball starts at height 10.5 metres at $t=0$. - It reaches maximum height 46.5 metres at $t=3$ seconds. - It hits the ground at $t=6.41$ seconds. - The horizontal line $h=16$ intersects the parabola at $t=0.24$ and $t=5.76$ seconds. This completes the solution to the problem.