A stone is dropped into a river from a bridge 53.6 m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone ? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released .

Answer

To solve the problem step by step, we will break it down into parts (a) and (b) as requested.

### Part (a): Finding the Initial Speed of the Second Stone

1. **Identify the parameters for the first stone:**
- Initial speed (u₁) = 0 m/s (since it is dropped)
- Acceleration (a) = -9.8 m/s² (downward, due to gravity)
- Displacement (s₁) = -53.6 m (downward)

2. **Use the equation of motion to find the time taken by the first stone to hit the water:**
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ -53.6 = 0 \cdot t + \frac{1}{2} (-9.8) t^2 \]
This simplifies to:
\[ -53.6 = -4.9 t^2 \]
Rearranging gives:
\[ t^2 = \frac{53.6}{4.9} \approx 10.92 \]
Taking the square root:
\[ t \approx 3.3 \text{ seconds} \]

3. **Determine the time available for the second stone:**
The second stone is thrown 1 second after the first stone is dropped, so:
\[ t' = t - 1 = 3.3 - 1 = 2.3 \text{ seconds} \]

4. **Identify the parameters for the second stone:**
- Initial speed (u₂) = ? (to be determined)
- Acceleration (a) = -9.8 m/s²
- Displacement (s₂) = -53.6 m (downward)
- Time (t') = 2.3 seconds

5. **Use the equation of motion for the second stone:**
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ -53.6 = u₂ \cdot 2.3 + \frac{1}{2} (-9.8) (2.3)^2 \]
Calculating the second term:
\[ -53.6 = u₂ \cdot 2.3 - 25.47 \]
Rearranging gives:
\[ u₂ \cdot 2.3 = -53.6 + 25.47 \]
\[ u₂ \cdot 2.3 = -28.13 \]
Solving for u₂:
\[ u₂ = \frac{-28.13}{2.3} \approx -12.03 \text{ m/s} \]
The negative sign indicates that the stone is thrown downward.

6. **Final answer for part (a):**
The initial speed of the second stone is approximately **12.03 m/s downward**.

### Part (b): Plotting Velocity vs. Time Graph

1. **Velocity of the first stone:**
- The velocity of the first stone as a function of time is given by:
\[ v_1 = u_1 + at = 0 - 9.8t = -9.8t \]
- At \(t = 0\), \(v_1 = 0\) m/s.
- At \(t = 3.3\), \(v_1 = -9.8 \times 3.3 \approx -32.34\) m/s.

2. **Velocity of the second stone:**
- The velocity of the second stone as a function of time is given by:
\[ v_2 = u_2 + at = -12.03 - 9.8t \]
- The second stone starts moving at \(t = 1\) second.
- At \(t = 1\), \(v_2 = -12.03 - 9.8 \times 1 = -21.83\) m/s.
- At \(t = 3.3\), \(v_2 = -12.03 - 9.8 \times 2.3 \approx -32.34\) m/s.

3. **Plotting the graph:**
- For the first stone, plot the line starting from (0, 0) to (3.3, -32.34).
- For the second stone, plot the line starting from (1, -21.83) to (3.3, -32.34).