一个球被竖直向上抛出。在抛出 $t$ 秒后，它距离地面的高度 $h$ 米由公式 $h = 7 + 20t - 5t^{2}$ 给出。求该球达到的最大高度以及达到该高度的时间。 求常数 $k$ 的取值范围，使得曲线 $y = x^{2} + 6x + k$ 完全位于直线 $y = k(x - 1)$ 的上方。

1. Factorise the coefficient of $t^2$ from the $t$ terms

$h = -5(t^2 - 4t) + 7$

2. Complete the square for the expression inside the bracket

$h = -5[(t - 2)^2 - (-2)^2] + 7$
$h = -5[(t - 2)^2 - 4] + 7$

3. Expand and simplify to find the vertex form

$h = -5(t - 2)^2 + 20 + 7$
$h = -5(t - 2)^2 + 27$

4. Identify the maximum height and the time it occurs

Since $-5(t - 2)^2 \leqslant 0$ for all $t$, the maximum value of $h$ is $27$.
This occurs when $t - 2 = 0$, so $t = 2$.

1. Set up the inequality for the curve to lie entirely above the line

For the curve to lie entirely above the line, we must have:
$x^2 + 6x + k > k(x - 1)$ for all real $x$.

2. Rearrange the inequality into standard quadratic form

$x^2 + 6x + k - kx + k > 0$
$x^2 + (6 - k)x + 2k > 0$

3. Apply the discriminant condition for a quadratic to be always positive

Since the coefficient of $x^2$ is $1 > 0$, the quadratic is always positive if the discriminant is negative ($b^2 - 4ac < 0$):
$(6 - k)^2 - 4(1)(2k) < 0$
$36 - 12k + k^2 - 8k < 0$
$k^2 - 20k + 36 < 0$

4. Solve the quadratic inequality for $k$

$(k - 2)(k - 18) < 0$
$2 < k < 18$