The ball is dropped from a bridge ` 122.5 m` above a river, After the ball has been falling for 2 s, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time ?

Options

A. `40 ms^(-1)`
B. `55.5 ms^(-1)`
C. `26.1 ms^(-1)`
D. `9.6 ms^(-1)`

Explanation

To solve the problem step by step, we need to analyze the motion of both balls and determine the initial velocity required for the second ball to hit the water at the same time as the first ball.

### Step 1: Determine the time taken by the first ball to hit the water.

The first ball is dropped from a height of \( h = 122.5 \, \text{m} \). The equation of motion for the first ball is:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) is the distance fallen (122.5 m),
- \( u \) is the initial velocity (0 m/s, since it is dropped),
- \( a \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( t \) is the time taken to hit the water.

Substituting the values, we get:

\[ 122.5 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

This simplifies to:

\[ 122.5 = 4.9 t^2 \]

Now, solving for \( t^2 \):

\[ t^2 = \frac{122.5}{4.9} \approx 25 \]

Taking the square root:

\[ t = \sqrt{25} = 5 \, \text{s} \]

### Step 2: Determine the time taken by the second ball.

The second ball is thrown downwards 2 seconds after the first ball is dropped. Therefore, the time taken by the second ball to hit the water is:

\[ t' = t - 2 = 5 - 2 = 3 \, \text{s} \]

### Step 3: Set up the equation for the second ball.

Let \( u \) be the initial velocity of the second ball. The equation of motion for the second ball is:

\[ s = ut' + \frac{1}{2} a (t')^2 \]

Substituting the known values:

\[ 122.5 = u \cdot 3 + \frac{1}{2} \cdot 9.8 \cdot (3)^2 \]

Calculating \( \frac{1}{2} \cdot 9.8 \cdot 9 \):

\[ \frac{1}{2} \cdot 9.8 \cdot 9 = 44.1 \]

Now substituting back into the equation:

\[ 122.5 = 3u + 44.1 \]

### Step 4: Solve for \( u \).

Rearranging the equation to isolate \( u \):

\[ 3u = 122.5 - 44.1 \]

Calculating the right side:

\[ 3u = 78.4 \]

Now dividing by 3:

\[ u = \frac{78.4}{3} \approx 26.13 \, \text{m/s} \]

### Final Answer:

The initial velocity \( u \) required for the second ball to hit the water at the same time as the first ball is approximately:

\[ u \approx 26.1 \, \text{m/s} \]

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