MAT 132 TEST 2 Review (Jan 2025 ) 1. If a ball is to thrown verically upwards, the ball reaches a height of 10 fl a) $$ y=-4 t^{2}+6 t+13 $$ b) $$ y=-7 t^{2}+2(+23 $$ feet after $$ t $$ seconds. i) find max height ii) when ball reach Gack to ground iii) when does ball reaches $$ 12 \mathrm{ft}, 5 \mathrm{ft} $$

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MAT 132 TEST 2 Review (Jan 2025 ) 1. If a ball is to thrown verically upwards, the ball reaches a height of 10 fl a) $$ y=-4 t^{2}+6 t+13 $$ b) $$ y=-7 t^{2}+2(+23 $$ feet after $$ t $$ seconds. 	i) find max height 	ii) when ball reach Gack to ground 	iii) when does ball reaches $$ 12 \mathrm{ft}, 5 \mathrm{ft} $$

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To solve the problem, we will analyze the two equations given for the height of the ball as a function of time $$ t $$.

### Given Equations:
1. $$ y = -4t^2 + 6t + 13 $$
2. $$ y = -7t^2 + 2t + 23 $$

We will solve the questions for both equations separately.

### For Equation (a): $$ y = -4t^2 + 6t + 13 $$

#### i) Find the maximum height
The maximum height of a quadratic function $$ y = at^2 + bt + c $$ occurs at $$ t = -\frac{b}{2a} $$.

Here, $$ a = -4 $$ and $$ b = 6 $$.

Calculating $$ t $$:
$$
t = -\frac{6}{2 \cdot -4} = \frac{6}{8} = \frac{3}{4} \text{ seconds}
$$

Now, substituting $$ t = \frac{3}{4} $$ back into the equation to find the maximum height:
$$
y = -4\left(\frac{3}{4}\right)^2 + 6\left(\frac{3}{4}\right) + 13
$$

Calculating this step by step:
$$
y = -4 \cdot \frac{9}{16} + \frac{18}{4} + 13
$$
$$
y = -\frac{36}{16} + \frac{72}{16} + \frac{208}{16}
$$
$$
y = \frac{72 - 36 + 208}{16} = \frac{244}{16} = 15.25 \text{ feet}
$$

#### ii) When does the ball reach the ground?
To find when the ball reaches the ground, we set $$ y = 0 $$:
$$
-4t^2 + 6t + 13 = 0
$$

We will use the quadratic formula $$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:
$$
t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot -4 \cdot 13}}{2 \cdot -4}
$$
Calculating the discriminant:
$$
b^2 - 4ac = 36 + 208 = 244
$$
Now substituting back into the formula:
$$
t = \frac{-6 \pm \sqrt{244}}{-8}
$$

Calculating $$ \sqrt{244} $$:
$$
\sqrt{244} \approx 15.62
$$
Now substituting:
$$
t = \frac{-6 \pm 15.62}{-8}
$$

Calculating the two possible values for $$ t $$:
1. $$ t_1 = \frac{-6 + 15.62}{-8} $$
2. $$ t_2 = \frac{-6 - 15.62}{-8} $$

Calculating $$ t_2 $$:
$$
t_2 = \frac{-21.62}{-8} \approx 2.70 \text{ seconds}
$$

Calculating $$ t_1 $$:
$$
t_1 = \frac{9.62}{-8} \text{ (not valid since time cannot be negative)}
$$

Thus, the ball reaches the ground at approximately $$ t \approx 2.70 $$ seconds.

#### iii) When does the ball reach $$ 12 \text{ ft} $$ and $$ 5 \text{ ft} $$?
1. For $$ y = 12 $$:
$$
-4t^2 + 6t + 13 = 12
$$
$$
-4t^2 + 6t + 1 = 0
$$

Using the quadratic formula:
$$
t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot -4 \cdot 1}}{2 \cdot -4}
$$
Calculating the discriminant:
$$
b^2 - 4ac = 36 + 16 = 52
$$
Now substituting back into the formula:
$$
t = \frac{-6 \pm \sqrt{52}}{-8}
$$
Calculating $$ \sqrt{52} $$:
$$
\sqrt{52} \approx 7.21
$$
Now substituting:
$$
t = \frac{-6 \pm 7.21}{-8}
$$

Calculating the two possible values for $$ t $$:
1. $$ t_1 = \frac{1.21}{-8} \text{ (not valid)} $$
2. $$ t_2 = \frac{-13.21}{-8} \approx 1.65 \text{ seconds} $$

2. For $$ y = 5 $$:
$$
-4t^2 + 6t + 13 = 5
$$
$$
-4t^2 + 6t + 8 = 0
$$

Using the quadratic formula:
$$
t = \frac{-6 \pm \sqrt{6^2 - 4 \cdot -4 \cdot 8}}{2 \cdot -4}
$$
Calculating the discriminant:
$$
b^2 - 4ac = 36 + 128 = 164
$$
Now substituting back into the formula:
$$
t = \frac{-6 \pm \sqrt{164}}{-8}
$$
Calculating $$ \sqrt{164} $$:
$$
\sqrt{164} \approx 12.81
$$
Now substituting:
$$
t = \frac{-6 \pm 12.81}{-8}
$$

Calculating the two possible values for $$ t $$:
1. $$ t_1 = \frac{6.81}{-8} \text{ (not valid)} $$
2. $$ t_2 = \frac{-18.81}{-8} \approx 2.35 \text{ seconds} $$

### Summary of Results for Equation (a):
- Maximum height: $$ 15.25 \text{ ft} $$
- Time to reach the ground: $$ 2.70 \text{ seconds} $$
- Time to reach $$ 12 \text{ ft} $$: $$ 1.65 \text{ seconds} $$
- Time to reach $$ 5 \text{ ft} $$: $$ 2.35 \text{ seconds} $$

### Now, let's analyze Equation (b): $$ y = -7t^2 + 2t + 23 $$

#### i) Find the maximum height
Using the same method:
$$
t = -\frac{2}{2 \cdot -7} = \frac{2}{14} = \frac{1}{7} \text{ seconds}
$$

Substituting $$ t =
For equation (a): - Maximum height: 15.25 feet - Time to reach the ground: 2.70 seconds - Time to reach 12 ft: 1.65 seconds - Time to reach 5 ft: 2.35 seconds For equation (b): - Maximum height: $$ \frac{1}{7} $$ seconds - Time to reach the ground: - Time to reach 12 ft: - Time to reach 5 ft:

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