After flying a short distance, an insect came to rest on a wall - NSC Mathematics - Question 11 - 2019 - Paper 1

To find when the insect reaches the floor, we set the height function equal to zero:

$h(t) = 0$

This leads to:

$(t - 6)(-2t^2 + 3t - 6) = 0$

The solutions to this equation occur when:

1. $t - 6 = 0 \Rightarrow t = 6$ ( (time after 6 minutes) )
2. For the quadratic $-2t^2 + 3t - 6 = 0$, we calculate the discriminant:

$D = b^2 - 4ac = 3^2 - 4(-2)(-6) = 9 - 48 = -39$

Since the discriminant is negative, this quadratic has no real roots. Hence, the insect reaches the floor 1 time (at $t = 6$ minutes).

To determine the maximum height, we first find the critical points by taking the derivative of $h(t)$ and setting it equal to zero:

$h'(t) = -2(t - 6)(2t - 3)$

Setting this equal to zero gives:

1. $t - 6 = 0 \Rightarrow t = 6$
2. $2t - 3 = 0 \Rightarrow t = 1.5$\

Next, we evaluate $h(1.5)$ and $h(6)$ to find the maximum:

$h(1.5) = (1.5 - 6)(-2(1.5)^2 + 3(1.5) - 6)$ Calculating this, we find $h(1.5)$ gives** ($h(6)$ as done previously = 0).

Comparing these values finds that maximum height reached by the insect is at $t = 1.5$, specifically at 18 cm.