While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.20 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant ? Take the downward direction to be the negative direction .

Options

A. `-4.9 m//s`
B. `-9.8 m//s`
C. `-8.4 m//s`
D. `-11` m/s

Answer

To solve the problem step by step, we will analyze the motion of both stones and use the equations of motion to find the initial velocity needed for the second stone.

### Step 1: Understand the scenario
- We have a bridge that is 15.0 m above the ground.
- A stone is dropped from rest (initial velocity \( u_1 = 0 \)) and falls 3.20 m before the second stone is thrown.
- We need to find the initial velocity \( u_2 \) for the second stone so that both stones hit the ground at the same time.

### Step 2: Calculate the time taken for the first stone to fall 3.20 m
Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = -3.20 \, \text{m} \) (downward displacement)
- \( u = 0 \) (initial velocity)
- \( a = -9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ -3.20 = 0 \cdot t + \frac{1}{2} (-9.81) t^2 \]
\[ -3.20 = -4.905 t^2 \]
\[ t^2 = \frac{3.20}{4.905} \approx 0.651 \]
\[ t \approx \sqrt{0.651} \approx 0.81 \, \text{s} \]

### Step 3: Calculate the total time taken for the first stone to reach the ground
The total time \( T \) for the first stone to fall from the bridge to the ground (15.0 m) can be calculated using the same equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = -15.0 \, \text{m} \)
- \( u = 0 \)
- \( a = -9.81 \, \text{m/s}^2 \)

Substituting the values:
\[ -15.0 = 0 \cdot T + \frac{1}{2} (-9.81) T^2 \]
\[ -15.0 = -4.905 T^2 \]
\[ T^2 = \frac{15.0}{4.905} \approx 3.06 \]
\[ T \approx \sqrt{3.06} \approx 1.75 \, \text{s} \]

### Step 4: Calculate the time available for the second stone to fall
The second stone is thrown after the first stone has fallen for \( t \approx 0.81 \, \text{s} \). Therefore, the time left for the second stone to reach the ground is:
\[ T - t \approx 1.75 - 0.81 \approx 0.94 \, \text{s} \]

### Step 5: Calculate the distance the second stone must fall
The second stone must fall the remaining distance:
\[ \text{Remaining distance} = 15.0 - 3.20 = 11.80 \, \text{m} \]

### Step 6: Use the equation of motion for the second stone
For the second stone, we use:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s = -11.80 \, \text{m} \)
- \( u = u_2 \) (initial velocity we need to find)
- \( a = -9.81 \, \text{m/s}^2 \)
- \( t \approx 0.94 \, \text{s} \)

Substituting the values:
\[ -11.80 = u_2(0.94) + \frac{1}{2} (-9.81)(0.94)^2 \]
Calculating \( \frac{1}{2} (-9.81)(0.94)^2 \):
\[ = -4.905 \cdot 0.8836 \approx -4.34 \]
Now substituting back:
\[ -11.80 = 0.94 u_2 - 4.34 \]
\[ 0.94 u_2 = -11.80 + 4.34 \]
\[ 0.94 u_2 = -7.46 \]
\[ u_2 = \frac{-7.46}{0.94} \approx -7.93 \, \text{m/s} \]

### Final Answer
The initial velocity \( u_2 \) that must be given to the second stone is approximately **7.93 m/s downward**.