Answer :
We are given the height function
[tex]$$
h(t) = -4.9t^2 + 40t,
$$[/tex]
which represents the height of the ball (in meters) at time [tex]$t$[/tex] seconds. We need to find the average rate of change of [tex]$h(t)$[/tex] over the interval [tex]$[1,4]$[/tex].
Step 1. Evaluate the height at [tex]$t=1$[/tex]:
[tex]$$
h(1) = -4.9(1)^2 + 40(1) = -4.9 + 40 = 35.1 \text{ meters}.
$$[/tex]
Step 2. Evaluate the height at [tex]$t=4$[/tex]:
[tex]$$
h(4) = -4.9(4)^2 + 40(4) = -4.9(16) + 160 = -78.4 + 160 = 81.6 \text{ meters}.
$$[/tex]
Step 3. Compute the average rate of change. The formula for the average rate of change between [tex]$t=a$[/tex] and [tex]$t=b$[/tex] is
[tex]$$
\text{Average Rate} = \frac{h(b) - h(a)}{b - a}.
$$[/tex]
For the interval [tex]$[1,4]$[/tex], we substitute:
[tex]$$
\text{Average Rate} = \frac{h(4) - h(1)}{4 - 1} = \frac{81.6 - 35.1}{3} = \frac{46.5}{3} = 15.5 \text{ meters per second}.
$$[/tex]
This calculation shows that over the interval [tex]$[1,4]$[/tex], on average, the height of the ball increases by [tex]$15.5$[/tex] meters for each second that passes.
Thus, the answers to the drag and drop question are:
1. The average rate of change of the quadratic function over the interval [tex]$[1,4]$[/tex] is [tex]$\boxed{15.5}$[/tex].
2. This means that, on average, the height of the ball [tex]$\boxed{\text{increases by 15.5}}$[/tex] meters for every increase of 1 second over the interval [tex]$[1,4]$[/tex].
[tex]$$
h(t) = -4.9t^2 + 40t,
$$[/tex]
which represents the height of the ball (in meters) at time [tex]$t$[/tex] seconds. We need to find the average rate of change of [tex]$h(t)$[/tex] over the interval [tex]$[1,4]$[/tex].
Step 1. Evaluate the height at [tex]$t=1$[/tex]:
[tex]$$
h(1) = -4.9(1)^2 + 40(1) = -4.9 + 40 = 35.1 \text{ meters}.
$$[/tex]
Step 2. Evaluate the height at [tex]$t=4$[/tex]:
[tex]$$
h(4) = -4.9(4)^2 + 40(4) = -4.9(16) + 160 = -78.4 + 160 = 81.6 \text{ meters}.
$$[/tex]
Step 3. Compute the average rate of change. The formula for the average rate of change between [tex]$t=a$[/tex] and [tex]$t=b$[/tex] is
[tex]$$
\text{Average Rate} = \frac{h(b) - h(a)}{b - a}.
$$[/tex]
For the interval [tex]$[1,4]$[/tex], we substitute:
[tex]$$
\text{Average Rate} = \frac{h(4) - h(1)}{4 - 1} = \frac{81.6 - 35.1}{3} = \frac{46.5}{3} = 15.5 \text{ meters per second}.
$$[/tex]
This calculation shows that over the interval [tex]$[1,4]$[/tex], on average, the height of the ball increases by [tex]$15.5$[/tex] meters for each second that passes.
Thus, the answers to the drag and drop question are:
1. The average rate of change of the quadratic function over the interval [tex]$[1,4]$[/tex] is [tex]$\boxed{15.5}$[/tex].
2. This means that, on average, the height of the ball [tex]$\boxed{\text{increases by 15.5}}$[/tex] meters for every increase of 1 second over the interval [tex]$[1,4]$[/tex].