Math  /  Algebra

Question15. A science experiment involves launching a small röcket. The following measurements are taken: Initial height: 0.61 m Initial vertical velocity: 36.85 m/s36.85 \mathrm{~m} / \mathrm{s} a) Create a quadratic model for the height, in metres, of the rocket after a given number of seconds. b) Verify the following results of the experiment: Total time in the air: 7.54 s Maximum height: 69.89 m c) Sketch a graph of this relation and label the key information as in Example 1 of this section.

Studdy Solution

STEP 1

What is this asking? We need to figure out the height of a rocket over time, given its starting height and speed, check some experimental results, and then draw a graph of the rocket's journey. Watch out! Gravity is a downwards acceleration, so be careful with the sign!

STEP 2

1. Define the function
2. Calculate total time in the air
3. Calculate maximum height
4. Sketch the graph

STEP 3

Alright, let's **define** our height function!
We know the general formula for height under constant acceleration is h(t)=12at2+v0t+h0 h(t) = \frac{1}{2}at^2 + v_0t + h_0 , where h(t)h(t) is the height at time tt, aa is the acceleration due to gravity, v0v_0 is the **initial velocity**, and h0h_0 is the **initial height**.

STEP 4

On Earth, the acceleration due to gravity is approximately 9.8 m/s2-9.8 \text{ m/s}^2.
Notice the negative sign – super important, because gravity pulls *downwards*!
Our **initial velocity** is 36.8536.85 m/s, and the **initial height** is 0.610.61 m.

STEP 5

Plugging these values into our formula, we get: h(t)=12(9.8)t2+36.85t+0.61 h(t) = \frac{1}{2} \cdot (-9.8) \cdot t^2 + 36.85t + 0.61 h(t)=4.9t2+36.85t+0.61 h(t) = -4.9t^2 + 36.85t + 0.61 This is our **quadratic model** for the rocket's height!

STEP 6

The rocket lands when h(t)=0h(t) = 0.
So, we need to solve 0=4.9t2+36.85t+0.610 = -4.9t^2 + 36.85t + 0.61.

STEP 7

We can use the **quadratic formula**: t=b±b24ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=4.9a = -4.9, b=36.85b = 36.85, and c=0.61c = 0.61.

STEP 8

Substituting these values, we get: t=36.85±36.8524(4.9)0.612(4.9) t = \frac{-36.85 \pm \sqrt{36.85^2 - 4 \cdot (-4.9) \cdot 0.61}}{2 \cdot (-4.9)} t=36.85±1357.9225+11.9569.8 t = \frac{-36.85 \pm \sqrt{1357.9225 + 11.956}}{-9.8} t=36.85±1369.87859.8 t = \frac{-36.85 \pm \sqrt{1369.8785}}{-9.8} t=36.85±37.0118699.8 t = \frac{-36.85 \pm 37.011869}{-9.8}

STEP 9

This gives us two possible values for tt: t=36.85+37.0118699.80.0165 t = \frac{-36.85 + 37.011869}{-9.8} \approx -0.0165 and t=36.8537.0118699.87.54 t = \frac{-36.85 - 37.011869}{-9.8} \approx 7.54

STEP 10

Since time can't be negative, the rocket is in the air for approximately **7.54 seconds**.
This matches the experimental result!

STEP 11

The maximum height occurs at the vertex of the parabola.
The time at which this occurs is given by t=b2at = -\frac{b}{2a}.

STEP 12

Using our values of aa and bb, we get: t=36.852(4.9)3.76 t = -\frac{36.85}{2 \cdot (-4.9)} \approx 3.76

STEP 13

Now, we plug this value of tt back into our height function: h(3.76)=4.9(3.76)2+36.85(3.76)+0.61 h(3.76) = -4.9(3.76)^2 + 36.85(3.76) + 0.61 h(3.76)=4.9(14.1376)+138.626+0.61 h(3.76) = -4.9(14.1376) + 138.626 + 0.61 h(3.76)=69.27424+138.626+0.61 h(3.76) = -69.27424 + 138.626 + 0.61 h(3.76)69.96 h(3.76) \approx 69.96

STEP 14

So, the **maximum height** is approximately **69.96 m**.
This is very close to the experimental result!

STEP 15

We'll sketch a parabola with the following key points: * **Starting point:** (0,0.61)(0, 0.61) * **Landing point:** (7.54,0)(7.54, 0) * **Vertex (maximum height):** (3.76,69.96)(3.76, 69.96)

STEP 16

The parabola should open downwards because the coefficient of the t2t^2 term is negative.
Label the axes as *time* (tt) and *height* (h(t)h(t)).

STEP 17

The rocket's height is modeled by h(t)=4.9t2+36.85t+0.61h(t) = -4.9t^2 + 36.85t + 0.61.
It stays in the air for **7.54 seconds**, reaching a **maximum height** of **69.96 m**.
The graph is a downward-opening parabola with the key points identified above.

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