A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?

Options

A. 2.45 m/s
B. 49 m/s
C. 14.5 m/s
D. 24.5 m/s

Explanation

To solve the problem, we need to determine the initial velocity of the second ball so that both balls hit the water at the same time. Let's break down the solution step by step.

### Step 1: Analyze the motion of the first ball
The first ball is dropped from a height of 176.4 m. The distance it travels after falling for 2 seconds can be calculated using the second equation of motion:

\[ h_1 = \frac{1}{2} g t^2 \]

where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)) and \( t \) is the time (2 seconds).

Calculating \( h_1 \):

\[ h_1 = \frac{1}{2} \times 9.8 \times (2^2) = \frac{1}{2} \times 9.8 \times 4 = 19.6 \, \text{m} \]

### Step 2: Calculate the remaining height for the first ball
After 2 seconds, the height of the first ball above the water is:

\[ h_{\text{remaining}} = 176.4 - 19.6 = 156.8 \, \text{m} \]

### Step 3: Determine the time taken for the first ball to hit the water
Let \( t_0 \) be the time taken for the first ball to hit the water after the initial 2 seconds. The total time for the first ball to reach the water is \( t_0 + 2 \) seconds. We can use the second equation of motion again:

\[ h_{\text{remaining}} = \frac{1}{2} g t_0^2 \]

Substituting \( h_{\text{remaining}} \):

\[ 156.8 = \frac{1}{2} \times 9.8 \times t_0^2 \]

Solving for \( t_0^2 \):

\[ t_0^2 = \frac{156.8 \times 2}{9.8} = \frac{313.6}{9.8} \approx 32 \]

Taking the square root:

\[ t_0 \approx 4 \, \text{s} \]

### Step 4: Analyze the motion of the second ball
The second ball is thrown downward 2 seconds after the first ball. It will also take \( t_0 \) seconds to hit the water. The total time for the second ball from the moment it is thrown until it hits the water is \( t_0 \).

Using the second equation of motion for the second ball:

\[ h_{\text{remaining}} = v_0 t_0 + \frac{1}{2} g t_0^2 \]

Substituting \( h_{\text{remaining}} \):

\[ 156.8 = v_0 \cdot 4 + \frac{1}{2} \times 9.8 \times (4^2) \]

Calculating the second term:

\[ \frac{1}{2} \times 9.8 \times 16 = 78.4 \]

Thus, we have:

\[ 156.8 = 4v_0 + 78.4 \]

### Step 5: Solve for the initial velocity \( v_0 \)
Rearranging the equation:

\[ 4v_0 = 156.8 - 78.4 \]

Calculating the right side:

\[ 4v_0 = 78.4 \]

Dividing by 4:

\[ v_0 = \frac{78.4}{4} = 19.6 \, \text{m/s} \]

### Final Answer
The initial velocity of the second ball should be \( 19.6 \, \text{m/s} \).

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