.)The following function gives the height, it metres,of a batted baseball as a function of the time, I seconds, since the ball was hit. h=-6(t-2.5)^2+38.5 a) What was the maximum height of the ball? b) What was the height of the ball when it was hit? c) How many seconds after it was hit did the ball hit the ground, to the nearest second? d) Find the height of the ball I after it was hit.

## Step 1: Finding the Maximum Height<br />### The maximum height of the ball occurs at the vertex of the parabola represented by the function $h(t) = -6(t-2.5)^2 + 38.5$. The vertex form of a parabola is $h(t) = a(t-h)^2 + k$, where $(h, k)$ is the vertex. In this case, the vertex is $(2.5, 38.5)$. The maximum height is the $k$ value of the vertex.<br /><br />## Step 2: Finding the Initial Height<br />### The height of the ball when it was hit corresponds to the height at $t=0$. Substitute $t=0$ into the equation: $h(0) = -6(0-2.5)^2 + 38.5 = -6(6.25) + 38.5 = -37.5 + 38.5 = 1$.<br /><br />## Step 3: Finding the Time to Hit the Ground<br />### The ball hits the ground when $h(t) = 0$. We need to solve the equation $-6(t-2.5)^2 + 38.5 = 0$.<br />### $-6(t-2.5)^2 = -38.5$<br />### $(t-2.5)^2 = \frac{38.5}{6} = \frac{77}{12}$<br />### $t-2.5 = \pm\sqrt{\frac{77}{12}}$<br />### $t = 2.5 \pm \sqrt{\frac{77}{12}}$<br />### Since time must be positive, we take the positive root: $t \approx 2.5 + \sqrt{6.4167} \approx 2.5 + 2.533 \approx 5.033$. Rounding to the nearest second gives $t=5$ seconds.<br /><br />## Step 4: Finding the Height 1 Second After Being Hit<br />### To find the height 1 second after being hit, substitute $t=1$ into the equation: $h(1) = -6(1-2.5)^2 + 38.5 = -6(-1.5)^2 + 38.5 = -6(2.25) + 38.5 = -13.5 + 38.5 = 25$.