Question

A water balloon is catapulted into the air so that its height h, in metres,
after t seconds is h =- 4.9t2+27t+2.4

a) How high is the balloon after 1 second?

b) For how long is the balloon more than 30 m high?

Answer 1

To solve the problem, let's break it down into the two parts as stated:

**Given the height function**:
\[ h(t) = -4.9t^2 + 27t + 2.4 \]

### a) How high is the balloon after 1 second?

We need to find the height at \( t = 1 \) second by plugging in \( t = 1 \) into the height function:

\[
h(1) = -4.9(1)^2 + 27(1) + 2.4
\]

Calculating this step-by-step:

\[
h(1) = -4.9(1) + 27(1) + 2.4
\]
\[
h(1) = -4.9 + 27 + 2.4
\]
\[
h(1) = 24.5 \text{ metres}
\]

So, the balloon is **24.5 metres** high after 1 second.

---

### b) For how long is the balloon more than 30 m high?

We need to find the values of \( t \) for which \( h(t) > 30 \). We can set up the inequality:

\[
-4.9t^2 + 27t + 2.4 > 30
\]

To solve this, first move all terms to one side:

\[
-4.9t^2 + 27t + 2.4 - 30 > 0
\]
\[
-4.9t^2 + 27t - 27.6 > 0
\]

Next, we can multiply the entire inequality by \(-1\) (remember that this reverses the inequality):

\[
4.9t^2 - 27t + 27.6 < 0
\]

Now, we will use the quadratic formula to find the roots of the equation \( 4.9t^2 - 27t + 27.6 = 0 \):

The quadratic formula is:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 4.9 \), \( b = -27 \), and \( c = 27.6 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-27)^2 - 4 \cdot 4.9 \cdot 27.6
\]
\[
= 729 - (4 \cdot 4.9 \cdot 27.6)
\]
\[
= 729 - 540.48 = 188.52
\]

Now we can find the roots:

\[
t = \frac{27 \pm \sqrt{188.52}}{2 \cdot 4.9}
\]

Calculating \( \sqrt{188.52} \):

\[
\sqrt{188.52} \approx 13.72
\]

Now, substituting back into the formula gives:

\[
t = \frac{27 \pm 13.72}{9.8}
\]

Calculating the two roots:

1.
\[
t_1 = \frac{27 + 13.72}{9.8} \approx \frac{40.72}{9.8} \approx 4.15
\]

2.
\[
t_2 = \frac{27 - 13.72}{9.8} \approx \frac{13.28}{9.8} \approx 1.36
\]

So, the roots are approximately \( t_1 \approx 4.15 \) seconds and \( t_2 \approx 1.36 \) seconds.

The balloon is above 30 metres between these two times:
\[
1.36 < t < 4.15
\]

So, the balloon is more than 30 metres high for approximately **2.79 seconds** (from about 1.36 seconds to about 4.15 seconds).