(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.

Certainly! Let's tackle each part step by step. --- ### **(i) How much higher than Harry's ball is Gina's ball half a second after Gina drops her ball (\( t = 1.5 \))?** **Given:** - Harry's ball height: \( h(t) = 1 + 18t - 5t^2 \) - Gina's ball height: \( g(t) = 18 + 10t - 5t^2 \) - Time after Gina drops her ball: \( t = 1.5 \) seconds **Objective:** Find the difference in height between Gina's ball and Harry's ball at \( t = 1.5 \) seconds, i.e., \( g(t) - h(t) \). **Step-by-Step Solution:** 1. **Calculate \( g(t) - h(t) \):** \[ g(t) - h(t) = (18 + 10t - 5t^2) - (1 + 18t - 5t^2) \] 2. **Simplify the expression:** \[ g(t) - h(t) = 18 + 10t - 5t^2 - 1 - 18t + 5t^2 \] \[ g(t) - h(t) = (18 - 1) + (10t - 18t) + (-5t^2 + 5t^2) \] \[ g(t) - h(t) = 17 - 8t \] 3. **Plug in \( t = 1.5 \) seconds:** \[ g(1.5) - h(1.5) = 17 - 8(1.5) = 17 - 12 = 5 \] **Conclusion:** Gina's ball is **5 units higher** than Harry's ball \( \frac{3}{2} \) seconds after Gina drops her ball. --- ### **(ii) Use calculus to find how long it takes Harry's ball to reach its maximum height.** **Given:** - Harry's ball height: \( h(t) = 1 + 18t - 5t^2 \) **Objective:** Determine the time \( t \) at which Harry's ball reaches its maximum height. **Step-by-Step Solution:** 1. **Find the derivative of \( h(t) \) with respect to time \( t \):** \[ h(t) = 1 + 18t - 5t^2 \] \[ h'(t) = \frac{d}{dt} [1] + \frac{d}{dt} [18t] - \frac{d}{dt} [5t^2] = 0 + 18 - 10t = 18 - 10t \] 2. **Set the derivative equal to zero to find critical points:** \[ h'(t) = 0 \Rightarrow 18 - 10t = 0 \] 3. **Solve for \( t \):** \[ 18 - 10t = 0 \Rightarrow 10t = 18 \Rightarrow t = \frac{18}{10} = 1.8 \text{ seconds} \] 4. **Verify that this critical point is a maximum:** - Since the coefficient of \( t^2 \) in \( h(t) \) is negative (\( -5 \)), the parabola opens downward, indicating that the critical point at \( t = 1.8 \) seconds is indeed a maximum. **Conclusion:** Harry's ball reaches its **maximum height after 1.8 seconds**. ---