Question

the takakaw waterfall in british columbia drops 302 metres in 4 distinct steps. if an object is thrown straight down from the tallest drop with an initial velocity of 10 m/s the height of the object above the ground can be modelled by the function d(t)=-5(t + 1)^2+265 where d(t) in metres is the height of the object, t seconds is the time for which the object has been falling. a) sketch a graph of the function. b) what part of the graph describes the falling object? explain. c) what is the height of the tallest drop of the falls in metres?

Explanation:

Step1: Analyze the function form

The function $d(t)=- 5(t + 1)^2+265$ is a quadratic - function in vertex form $y=a(x - h)^2+k$, where $a=-5$, $h=-1$, and $k = 265$. The graph of a quadratic function $y = ax^2+bx + c$ ($a
eq0$) is a parabola. When $a\lt0$, the parabola opens downwards.

Step2: Sketch the graph

1. The vertex of the parabola is at the point $(h,k)$. For the function $d(t)=-5(t + 1)^2+265$, the vertex is at $(-1,265)$. But since $t$ represents time, $t\geq0$.
2. When $t = 0$, $d(0)=-5(0 + 1)^2+265=-5+265 = 260$.
3. As $t$ increases, the value of $d(t)$ decreases because $a=-5\lt0$. We can plot the point $(0,260)$ and know the general shape of the parabola opening downwards for $t\geq0$.

Step3: Determine the falling - part of the graph

The object is falling when the height $d(t)$ is decreasing. Since the parabola $d(t)=-5(t + 1)^2+265$ ($a=-5\lt0$) opens downwards, for $t\geq0$, the entire part of the graph for $t\geq0$ describes the falling object. As time $t$ increases from $t = 0$, the height of the object is decreasing.

Step4: Find the height of the tallest drop

The object starts at $t = 0$ with height $d(0)=260$ meters and hits the ground when $d(t)=0$.
Set $d(t)=0$:

$$
LATEXBLOCK0
$$

The initial height is $d(0)=260$ meters and the final height is $d(6.2)=0$ meters. The height of the tallest drop is the initial height when $t = 0$, which is $260$ meters.

Answer:

a) To sketch the graph of $d(t)=-5(t + 1)^2+265$ for $t\geq0$: Plot the vertex at $(-1,265)$ (but consider only $t\geq0$), find $d(0)=260$, and draw a downward - opening parabola starting from the point $(0,260)$.
b) The part of the graph for $t\geq0$ describes the falling object because the quadratic function $d(t)$ with $a=-5\lt0$ is decreasing for $t\geq0$.
c) The height of the tallest drop is $260$ meters.