If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 4.5 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is $$ -32 \mathrm{ft} / \mathrm{sec}^{2} $$. [? ] feet

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UpStudy AI · Accepted Answer

To find the maximum height reached by the ball, we can use the kinematic equation for vertical motion:

$$
h(t) = h_0 + v_0 t + \frac{1}{2} a t^2
$$

where:
- $$ h(t) $$ is the height at time $$ t $$,
- $$ h_0 $$ is the initial height,
- $$ v_0 $$ is the initial velocity,
- $$ a $$ is the acceleration due to gravity,
- $$ t $$ is the time in seconds.

### Step 1: Identify Known Values
- Initial height, $$ h_0 = 4 $$ feet
- Acceleration due to gravity, $$ a = -32 $$ ft/s²
- Total time in the air, $$ t = 4.5 $$ seconds

### Step 2: Find the Initial Velocity
Since the ball is thrown upwards and then comes back down, it will take half of the total time to reach the maximum height. Therefore, the time to reach maximum height is:

$$
t_{up} = \frac{t}{2} = \frac{4.5}{2} = 2.25 \text{ seconds}
$$

At the maximum height, the final velocity is 0. We can use the following equation to find the initial velocity $$ v_0 $$:

$$
v = v_0 + at
$$

Setting $$ v = 0 $$ at the maximum height:

$$
0 = v_0 - 32 \cdot 2.25
$$

Now, we can solve for $$ v_0 $$:

$$
v_0 = 32 \cdot 2.25
$$

### Step 3: Calculate Initial Velocity
Let's calculate $$ v_0 $$:

$$
v_0 = 32 \cdot 2.25
$$

### Step 4: Calculate Maximum Height
Now we can substitute $$ v_0 $$ back into the height equation to find the maximum height:

$$
h_{max} = h_0 + v_0 t_{up} + \frac{1}{2} a t_{up}^2
$$

Substituting the known values:

$$
h_{max} = 4 + v_0 \cdot 2.25 + \frac{1}{2} \cdot (-32) \cdot (2.25)^2
$$

Now, let's perform the calculations.

I'll calculate $$ v_0 $$ and then $$ h_{max} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$32\times 2.25$$
- step1: Multiply the numbers:
  $$72$$
Calculate or simplify the expression $$ 4 + (32 * 2.25 * 2.25) + (0.5 * -32 * (2.25)^2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$4+\left(32\times 2.25\times 2.25\right)+\left(0.5\left(-32\right)\times 2.25^{2}\right)$$
- step1: Multiply the terms:
  $$4+162+\left(0.5\left(-32\right)\times 2.25^{2}\right)$$
- step2: Convert the expressions:
  $$4+162+\left(0.5\left(-32\right)\left(\frac{9}{4}\right)^{2}\right)$$
- step3: Multiply:
  $$4+162+\left(-81\right)$$
- step4: Remove the parentheses:
  $$4+162-81$$
- step5: Calculate:
  $$85$$
### Step 3: Calculate Initial Velocity
The initial velocity $$ v_0 $$ is calculated as:

$$
v_0 = 32 \cdot 2.25 = 72 \text{ ft/s}
$$

### Step 4: Calculate Maximum Height
Now substituting $$ v_0 $$ back into the height equation:

$$
h_{max} = 4 + (72 \cdot 2.25) + \left(\frac{1}{2} \cdot -32 \cdot (2.25)^2\right)
$$

Calculating this gives:

$$
h_{max} = 4 + 72 \cdot 2.25 + \left(-16 \cdot 5.0625\right)
$$

Calculating the terms:

1. $$ 72 \cdot 2.25 = 162 $$
2. $$ -16 \cdot 5.0625 = -81 $$

Now substituting these values back:

$$
h_{max} = 4 + 162 - 81 = 85 \text{ feet}
$$

### Final Answer
The maximum height reached by the ball is approximately **85 feet**.
The ball reached a maximum height of 85 feet.

If a ball is thrown in the air with an initial height of 4 feet, and if the ball remains in the air for 4.5 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is \( -32 \mathrm{ft} / \mathrm{sec}^{2} \). [? ] feet

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