A ball is dropped from the top of a tower of height 78.4 m Another ball is thrown down with a certain velocity 2 sec later. If both the balls reach the ground simultaneously, the velocity of the second ball is

Options

A. 29.4 m/s
B. 26.4 m/s
C. 30 m/s
D. 41.6 m/s

Answer

To solve the problem, we need to find the velocity of the second ball that is thrown downwards after 2 seconds, given that both balls reach the ground simultaneously.

### Step-by-Step Solution:

1. **Identify the height of the tower**:
The height of the tower is given as \( h = 78.4 \, \text{m} \).

2. **Calculate the time taken for the first ball to reach the ground**:
The first ball is dropped from the top of the tower. We can use the formula for the time of flight for an object under free fall:
\[ t_1 = \sqrt{\frac{2h}{g}} \]
where \( g \) (acceleration due to gravity) is approximately \( 9.8 \, \text{m/s}^2 \).
Substituting the values:
\[ t_1 = \sqrt{\frac{2 \times 78.4}{9.8}} = \sqrt{\frac{156.8}{9.8}} = \sqrt{16} = 4 \, \text{s} \]

3. **Determine the time taken for the second ball**:
The second ball is thrown downwards 2 seconds after the first ball. Therefore, if the first ball takes 4 seconds to reach the ground, the second ball will take:
\[ t_2 = t_1 - 2 = 4 - 2 = 2 \, \text{s} \]

4. **Use the second equation of motion for the second ball**:
The second ball is thrown downwards with an initial velocity \( u \) and travels the same height \( h = 78.4 \, \text{m} \) in \( t_2 = 2 \, \text{s} \). The second equation of motion is:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = 78.4 \, \text{m} \), \( a = g = 9.8 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \).
Substituting the values:
\[ 78.4 = u \cdot 2 + \frac{1}{2} \cdot 9.8 \cdot (2^2) \]
Simplifying the equation:
\[ 78.4 = 2u + \frac{1}{2} \cdot 9.8 \cdot 4 \]
\[ 78.4 = 2u + 19.6 \]

5. **Solve for \( u \)**:
Rearranging the equation to isolate \( u \):
\[ 2u = 78.4 - 19.6 \]
\[ 2u = 58.8 \]
\[ u = \frac{58.8}{2} = 29.4 \, \text{m/s} \]

### Final Answer:
The velocity of the second ball is \( 29.4 \, \text{m/s} \).