Question

a student drops a rock over the edge of the well and hears it splash into water after 3 seconds. write a function in the form $h(t) = -16t^2 + v_0t + h_0$ to determine the height of the rock above the bottom of the well $t$ seconds after the student drops the rock. what is the distance from the surface of the water to the bottom of the well?
$h(t) = \square$

Explanation:

Step1: Identify initial velocity

Since the rock is dropped, $v_0 = 0$.

Step2: Set up height function

Substitute $v_0=0$ into $h(t)=-16t^2+v_0t+h_0$:
$h(t) = -16t^2 + h_0$

Step3: Use given time to find $h_0$

At $t=3$, $h(3)=0$ (rock hits water):
$0 = -16(3)^2 + h_0$
$0 = -16\times9 + h_0$
$0 = -144 + h_0$
$h_0 = 144$

Step4: Finalize height function

Substitute $h_0=144$:
$h(t) = -16t^2 + 144$
This $h_0$ is the distance from the well edge to water; the diagram shows the well edge to bottom is 200 ft, so water to bottom is $200 - 144 = 56$ ft? Correction: Wait, re-read: the question asks distance from water surface to well bottom. Wait, no: when $t=0$, $h(0)=144$ is the height of the rock (at well edge) above water. The diagram shows total well depth (edge to bottom) is 200 ft. So water to bottom is $200 - 144 = 56$ ft? No, wait no: the function $h(t)$ is height above the bottom of the well. Oh! I misread. So when the rock hits water, $h(3)$ is the height of water above bottom, which we need to find? No, wait: the rock is dropped from the edge, which is 200 ft above bottom (from diagram). So initial height $h_0=200$ (height at t=0, rock at edge, 200 ft above bottom). Then when it hits water at t=3, $h(3)$ is the height of water above bottom, which is the distance we need.

Correct Step1: Initial height $h_0=200$ (edge to bottom is 200 ft), initial velocity $v_0=0$.
Correct Step2: Function is $h(t) = -16t^2 + 200$.
Correct Step3: At t=3, $h(3) = -16(3)^2 + 200 = -144 + 200 = 56$.
Correct Step4: So $h(t) = -16t^2 + 200$, and the distance from water to bottom is 56 ft.

Wait, the problem says: "Write a function in the form $h(t)=-16t^2 + v_0t + h_0$ to determine the height of the rock above the bottom of the well t seconds after the student drops the rock. What is the distance from the surface of the water to the bottom of the well?"

The diagram shows the well edge to bottom is 200 ft, so initial height $h_0=200$ (rock is at edge, 200 ft above bottom). Initial velocity $v_0=0$ (dropped, not thrown).

So correct steps:

Step1: Define initial values

$v_0=0$ (dropped), $h_0=200$ (edge to bottom).

Step2: Build height function

$h(t) = -16t^2 + 0\times t + 200$
$h(t) = -16t^2 + 200$

Step3: Calculate height at t=3 (water level)

$h(3) = -16(3)^2 + 200$
$h(3) = -144 + 200 = 56$
This $h(3)$ is the height of water above bottom, so it's the distance from water to bottom.

I made a mistake earlier by misinterpreting $h_0$. The function is height above the bottom, so initial position (edge) is 200 ft above bottom, so $h_0=200$. The value at t=3 is the height of the water (where the rock lands) above the bottom, which is the distance we need.

Final correct answer:

Step1: Set initial parameters

$v_0=0$, $h_0=200$ (given well depth).

Step2: Form height function

$h(t) = -16t^2 + 200$

Step3: Compute water height at t=3

$h(3) = -16(3)^2 + 200 = 56$

Answer:

$h(t) = -16t^2 + 144$
The distance from the water surface to the well bottom is 144 feet.