From the top of the tower, a ball `A` is dropped and another ball `B` is thrown horizontally at the same time. Which ball strikes the ground first ?

Options

A. `A`
B. `B`
C. simultaneously
D. depends on the masses

Explanation

To determine which ball strikes the ground first, we need to analyze the motion of both balls separately.

### Step-by-Step Solution:

1. **Understanding the Motion of Ball A:**
- Ball A is dropped from the top of the tower. This means it has an initial vertical velocity of \( u_{yA} = 0 \) m/s (since it is dropped).
- The only force acting on it is gravity, which accelerates it downwards at \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)).
- The vertical distance it falls is \( h \).

2. **Using the Equation of Motion:**
- We can use the second equation of motion:
\[ h = u_{yA} t_1 + \frac{1}{2} g t_1^2 \]
- Since \( u_{yA} = 0 \), this simplifies to:
\[ h = \frac{1}{2} g t_1^2 \]
- Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \]
- Taking the square root:
\[ t_1 = \sqrt{\frac{2h}{g}} \]

3. **Understanding the Motion of Ball B:**
- Ball B is thrown horizontally with an initial horizontal velocity \( u_{xB} = u \) and an initial vertical velocity \( u_{yB} = 0 \).
- The vertical motion is still influenced by gravity, so it also falls a distance \( h \).

4. **Using the Equation of Motion for Ball B:**
- The vertical motion for Ball B can be described by the same equation:
\[ h = u_{yB} t_2 + \frac{1}{2} g t_2^2 \]
- Since \( u_{yB} = 0 \):
\[ h = \frac{1}{2} g t_2^2 \]
- Rearranging gives:
\[ t_2^2 = \frac{2h}{g} \]
- Taking the square root:
\[ t_2 = \sqrt{\frac{2h}{g}} \]

5. **Comparing the Times:**
- From the calculations, we have:
\[ t_1 = \sqrt{\frac{2h}{g}} \quad \text{and} \quad t_2 = \sqrt{\frac{2h}{g}} \]
- Therefore, \( t_1 = t_2 \).

### Conclusion:
Both balls A and B strike the ground at the same time, as \( t_1 = t_2 \).