Worksheet featuring four quadratic equation word problems related to physics and business scenarios.

Let's solve each of these quadratic equation word problems step by step, explaining the reasoning and math behind each.

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Problem 1: Flare Launched from a Boat

> A flare is launched from a boat. The height, $ h $, in metres, of the flare above the water is approximately modelled by the function
> $$
> h(t) = -15t^2 + 150t
> $$
> where $ t $ is the number of seconds after the flare is launched. How many seconds will it take for the flare to return to the water?

#### Step 1: Understand the question
We want to know when the flare returns to the water, i.e., when its height $ h(t) = 0 $. So we solve:
$$
-15t^2 + 150t = 0
$$

#### Step 2: Solve the equation
Factor out common terms:
$$
-15t(t - 10) = 0
$$

Set each factor equal to zero:
- $ -15t = 0 $ → $ t = 0 $
- $ t - 10 = 0 $ → $ t = 10 $

#### Step 3: Interpret the solution
- $ t = 0 $: This is when the flare is launched.
- $ t = 10 $: This is when it returns to the water.

✔ Answer: It takes 10 seconds for the flare to return to the water.

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Problem 2: Osprey Diving for Salmon

> An osprey dives toward the water. Its height above the water $ t $ seconds after it begins its dive is given by:
> $$
> h(t) = 5t^2 - 30t + 45
> $$
> Determine the time it takes for the osprey to reach a height of 20 m.

#### Step 1: Set up the equation
We want $ h(t) = 20 $:
$$
5t^2 - 30t + 45 = 20
$$

#### Step 2: Simplify
Subtract 20 from both sides:
$$
5t^2 - 30t + 25 = 0
$$

Divide entire equation by 5:
$$
t^2 - 6t + 5 = 0
$$

#### Step 3: Factor
$$
(t - 1)(t - 5) = 0
$$

So:
- $ t = 1 $
- $ t = 5 $

#### Step 4: Interpret
The osprey reaches 20 meters at two times: once on the way down (but wait — let’s check direction).

But look at the quadratic: $ h(t) = 5t^2 - 30t + 45 $

This opens upward (positive coefficient of $ t^2 $), so the minimum height occurs at the vertex.

Wait! That suggests the osprey starts high, goes down, then comes back up? But that doesn't make sense — it's diving.

Let’s double-check the model.

Actually, if the osprey is diving downward, we expect the height to decrease over time. But this function has a positive leading coefficient, which means it opens upward, meaning it has a minimum point.

So the osprey starts at a height, dives down to a low point, then rises again? That contradicts reality unless it's not just a straight dive.

But the problem says “dives towards the water,” so likely it should be decreasing initially.

Let’s evaluate $ h(0) $:
$ h(0) = 45 $ m — starting height.

Now $ h(1) = 5(1)^2 - 30(1) + 45 = 5 - 30 + 45 = 20 $ m
$ h(2) = 5(4) - 60 + 45 = 20 - 60 + 45 = 5 $ m
$ h(3) = 5(9) - 90 + 45 = 45 - 90 + 45 = 0 $ m
$ h(4) = 5(16) - 120 + 45 = 80 - 120 + 45 = 5 $ m

Ah! So the osprey dives down, hits the water at $ t=3 $, then somehow goes back up? That doesn’t make sense physically.

Wait — maybe there's a typo in interpretation?

But the function is given as $ h(t) = 5t^2 - 30t + 45 $, which is valid per the problem.

But since the osprey dives, we assume it starts at $ t=0 $, $ h=45 $, and descends.

So the first time it reaches 20 m is at $ t=1 $, before reaching the lowest point.

At $ t=5 $, it's going back up — but that would require it to have hit the water and bounced, which isn't realistic.

But the problem doesn't say anything about bouncing — it just asks when it reaches 20 m.

So unless specified otherwise, both times are mathematically valid.

But physically, only the first time makes sense: $ t = 1 $ second.

But wait — does it reach 20 m on the way down?

Yes: at $ t = 1 $, it's descending from 45 m to minimum at $ t = 3 $ (since vertex is at $ t = \frac{30}{2 \cdot 5} = 3 $), so yes.

Then at $ t = 5 $, it's ascending — but that implies it went below water and came back up — impossible unless it's a different scenario.

But the model may be hypothetical.

So, mathematically, the osprey reaches 20 m at $ t = 1 $ and $ t = 5 $.

But physically, it likely reaches 20 m once on the way down.

So the answer is:

✔ Answer: It takes 1 second to reach 20 m (on the way down).
(Note: $ t = 5 $ is not physically meaningful unless the bird bounces, which is not implied.)

But since the problem doesn’t specify direction, we give both solutions.

But typically, "time to reach" implies first time.

So final answer: 1 second

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Problem 3: Baseball Popped Up

> Ted popped a baseball straight up with an initial upward velocity of 48 ft/s. The height $ h $ in feet is modeled by:
> $$
> h(t) = -16t^2 + 48t + 3
> $$
> How long was the ball in the air if the catcher catches the ball three feet above the ground? Is your answer reasonable?

#### Step 1: Understand the situation
- Ball is thrown up from 3 ft (initial height).
- It goes up, then comes down.
- Catcher catches it at 3 ft again.
- We want how long it was in the air — i.e., from launch ($ t = 0 $) to when it returns to $ h = 3 $ ft.

So set $ h(t) = 3 $:
$$
-16t^2 + 48t + 3 = 3
$$

Subtract 3:
$$
-16t^2 + 48t = 0
$$

Factor:
$$
-16t(t - 3) = 0
$$

Solutions:
- $ t = 0 $: launch time
- $ t = 3 $: catch time

#### Step 2: Answer
Time in air = $ 3 - 0 = 3 $ seconds.

#### Step 3: Reasonableness
Initial velocity: 48 ft/s
Acceleration due to gravity: ~32 ft/s² downward
Time to reach max height: $ t = \frac{v}{g} = \frac{48}{32} = 1.5 $ s
Then falls back down — symmetric motion if same height.

Since it starts and ends at 3 ft, the total time should be twice the time to peak: $ 2 \times 1.5 = 3 $ seconds — matches.

Also, max height:
$ h(1.5) = -16(2.25) + 48(1.5) + 3 = -36 + 72 + 3 = 39 $ ft — reasonable for a pop-up.

✔ Answer: The ball was in the air for 3 seconds. Yes, this is reasonable because the motion is symmetric, and the time matches physics expectations.

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Problem 4: Angela’s Surf Shop Profit

> Her profit is modeled by:
> $$
> P(t) = 1125(t - 1)^2 - 10,125
> $$
> where $ t $ is years of operation. Determine how long it will take for the shop to start making a profit.

#### Step 1: Understand what "start making a profit" means
Profit > 0 → $ P(t) > 0 $

So solve:
$$
1125(t - 1)^2 - 10,125 > 0
$$

But first, find when profit becomes zero (break-even point):
$$
1125(t - 1)^2 - 10,125 = 0
$$

Add 10,125:
$$
1125(t - 1)^2 = 10,125
$$

Divide both sides by 1125:
$$
(t - 1)^2 = \frac{10,125}{1125}
$$

Calculate:
$$
\frac{10,125}{1125} = 9
$$

So:
$$
(t - 1)^2 = 9
\Rightarrow t - 1 = \pm 3
\Rightarrow t = 1 + 3 = 4 \quad \text{or} \quad t = 1 - 3 = -2
$$

But $ t = -2 $ is not meaningful (before opening).

So break-even at $ t = 4 $

Now, since the parabola opens upward (coefficient of $ (t-1)^2 $ is positive), profit is negative before $ t = 4 $, zero at $ t = 4 $, and positive after.

So profit starts at $ t = 4 $.

But the question is: how long until it starts making a profit?

That’s after $ t = 4 $ — but technically, profit is zero at $ t = 4 $, and positive after.

So the shop starts making a profit after 4 years.

But often, "starts making a profit" means when profit becomes positive, so at $ t = 4 $.

But since it's exactly zero at $ t = 4 $, it's not yet profitable.

So we need to find when profit > 0.

From earlier:
$$
P(t) > 0 \quad \text{when} \quad (t - 1)^2 > 9 \Rightarrow |t - 1| > 3
\Rightarrow t > 4 \quad \text{or} \quad t < -2
$$

Only $ t > 4 $ is valid.

So profitable after 4 years.

But the earliest time it becomes profitable is just after $ t = 4 $.

So it takes 4 years to reach break-even, and after that it makes a profit.

Thus, it will take 4 years to start making a profit — assuming "start" includes break-even or the moment profit turns positive.

But strictly speaking, profit is zero at $ t = 4 $, so it starts making profit at $ t = 4 $.

But let's compute $ P(4) $:
$$
P(4) = 1125(4 - 1)^2 - 10,125 = 1125(9) - 10,125 = 10,125 - 10,125 = 0
$$

So no profit at $ t = 4 $.

Try $ t = 5 $:
$$
P(5) = 1125(4)^2 - 10,125 = 1125 \times 16 - 10,125 = 18,000 - 10,125 = 7,875 > 0
$$

So profit begins after 4 years.

Therefore, the shop starts making a profit after 4 years, meaning in year 5.

But the question is: how long will it take?

It takes 4 years to reach the break-even point, and only after that does profit begin.

So the shortest time it takes to start making a profit is after 4 years.

But since we can't say "after 4 years" precisely without a decimal, we might say:

✔ Answer: It will take more than 4 years to start making a profit. Specifically, at $ t = 4 $ it breaks even, and for $ t > 4 $ it makes a profit. So the shop starts making a profit after 4 years.

But if the question expects a specific value, it’s likely 4 years — the time when profit becomes non-negative.

However, strictly, profit starts at $ t = 4 $ only if we consider zero profit as "making profit", which is not standard.

In business, "making a profit" usually means positive profit.

So answer: After 4 years.

But to express it clearly:

> The shop will start making a profit after 4 years of operation.

But since the model gives exact break-even at $ t = 4 $, and profit increases after that, the minimum time is greater than 4 years.

But in most contexts, they expect:

✔ Answer: It will take 4 years for the shop to start making a profit.
(Though technically, it starts making profit after 4 years; however, $ t = 4 $ is the threshold.)

Alternatively, you could say:
The shop starts making a profit at $ t = 4 $ (if zero profit counts), or after $ t = 4 $ (if positive profit is required).

But since the problem says “start making a profit”, and profit is zero at $ t = 4 $, the best answer is:

> The shop will start making a profit after 4 years.

But if forced to give a number: 4 years is the answer expected.

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✔ Final Answers Summary:

1. 10 seconds
2. 1 second (first time reaching 20 m)
3. 3 seconds — reasonable due to symmetry and physics
4. After 4 years (or 4 years as the break-even point)

Let me know if you'd like diagrams or graphs for any of these!