If a ball is thrown in the air with an initial height of 3 feet, and if the ball remains in the air for 4 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 solve this problem, we can use the kinematic equation for the height of an object in free fall, which is given by:

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

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 (which is negative due to gravity),
- $$ t $$ is the time in seconds.

### Step 1: Identify the known values
- Initial height, $$ h_0 = 3 $$ feet
- Acceleration due to gravity, $$ a = -32 $$ ft/s²
- Total time in the air, $$ t = 4 $$ seconds

### Step 2: Determine the initial velocity
Since the ball is thrown upwards and remains in the air for 4 seconds, it will take 2 seconds to reach its maximum height (the time to go up is equal to the time to come down). We can use the following kinematic equation to find the initial velocity:

At the maximum height, the final velocity $$ v = 0 $$. Using the equation:

$$
v = v_0 + a t
$$

Setting $$ v = 0 $$ at $$ t = 2 $$ seconds (the time to reach maximum height):

$$
0 = v_0 - 32 \cdot 2
$$

Solving for $$ v_0 $$:

$$
v_0 = 32 \cdot 2 = 64 \text{ ft/s}
$$

### Step 3: Calculate the maximum height
Now we can find the maximum height using the time $$ t = 2 $$ seconds in the height equation:

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

Substituting the known values:

$$
h(2) = 3 + 64 \cdot 2 + \frac{1}{2} \cdot (-32) \cdot (2)^2
$$

Calculating each term:

1. $$ 64 \cdot 2 = 128 $$
2. $$ \frac{1}{2} \cdot (-32) \cdot 4 = -64 $$

Now substituting these values back into the equation:

$$
h(2) = 3 + 128 - 64
$$

Calculating the final height:

$$
h(2) = 3 + 128 - 64 = 67 \text{ feet}
$$

### Final Answer
The maximum height the ball reached is $$ \boxed{67} $$ feet.
The ball reached a maximum height of 67 feet.

If a ball is thrown in the air with an initial height of 3 feet, and if the ball remains in the air for 4 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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