A ball is thrown upward and reaches a height of `64` feet, its initial velocity should be `(g=32"ft/sec"^(2))`

Options

A. `64` ft/sec
B. `72` ft/sec
C. `32` ft/sec
D. `4096` ft/sec

Explanation

To solve the problem of finding the initial velocity of a ball thrown upward that reaches a height of 64 feet, we can use the kinematic equation:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity (0 ft/s at the highest point)
- \( u \) = initial velocity (what we are trying to find)
- \( a \) = acceleration (which is -g, where g = 32 ft/s²)
- \( s \) = displacement (64 feet)

### Step-by-Step Solution:

1. **Identify the known values:**
- Maximum height \( s = 64 \) feet
- Final velocity \( v = 0 \) ft/s (at the highest point)
- Acceleration due to gravity \( g = 32 \) ft/s² (acting downwards, hence \( a = -g = -32 \) ft/s²)

2. **Substitute the known values into the kinematic equation:**
\[ 0^2 = u^2 + 2(-32)(64) \]

3. **Simplify the equation:**
\[ 0 = u^2 - 64 \times 32 \]
\[ 0 = u^2 - 2048 \]

4. **Rearrange to solve for \( u^2 \):**
\[ u^2 = 2048 \]

5. **Take the square root to find \( u \):**
\[ u = \sqrt{2048} \]
\[ u = 32\sqrt{2} \text{ ft/s} \]

6. **Calculate the numerical value:**
\[ u \approx 32 \times 1.414 \approx 45.25 \text{ ft/s} \]

### Final Answer:
The initial velocity \( u \) should be approximately \( 45.25 \) ft/s.