(c) The height of Harry's ball during its fall, \( h(t) \), is given by: \[ h(t)=1+18 t-5 t^{2} \] Use this, and the fact that \( g(t)=18+10 t-5 t^{2} \), to answer the following questions. These are the same questions as in part (b), but here you will need to use algebra to answer them. (i) How much higher than Harry's ball is Gina's ball, half a second after Gina drops her ball ( \( t=1.5 \) )? Show your working out. (ii) Use calculus to find how long it takes Harry's ball to reach its maximum height.

To find the answers, let's break down the provided equations: For (i), we need to calculate the height of both Harry's ball \( h(t) \) and Gina's ball \( g(t) \) at \( t = 1.5 \) seconds. **Step 1: Calculate Harry's ball height at \( t = 1.5 \) seconds.** Plugging \( t = 1.5 \) into \( h(t) \): \[ h(1.5) = 1 + 18(1.5) - 5(1.5)^2 \] \[ = 1 + 27 - 5(2.25) \] \[ = 1 + 27 - 11.25 \] \[ = 16.75 \text{ meters} \] **Step 2: Calculate Gina's ball height at \( t = 1.5 \) seconds.** Plugging \( t = 1.5 \) into \( g(t) \): \[ g(1.5) = 18 + 10(1.5) - 5(1.5)^2 \] \[ = 18 + 15 - 5(2.25) \] \[ = 18 + 15 - 11.25 \] \[ = 21.75 \text{ meters} \] **Step 3: Determine the difference in height.** The height difference between Gina's ball and Harry's ball is: \[ g(1.5) - h(1.5) = 21.75 - 16.75 = 5 \text{ meters} \] So, half a second after Gina drops her ball, Gina’s ball is 5 meters higher than Harry’s ball. --- For (ii), to find the time when Harry's ball reaches its maximum height, we need to use calculus. The maximum height can be found at the vertex of the parabola defined by the quadratic equation \( h(t) \). The formula for the vertex \( t \) of a parabola given by \( ax^2 + bx + c \) is: \[ t = -\frac{b}{2a} \] In the equation \( h(t) = -5t^2 + 18t + 1 \), - \( a = -5 \) - \( b = 18 \) Plugging these values into the vertex formula: \[ t = -\frac{18}{2 \times -5} = \frac{18}{10} = 1.8 \text{ seconds} \] Thus, Harry's ball reaches its maximum height at \( t = 1.8 \) seconds.