A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that is 12 m from the point of impact when the key is released . What is the speed of the boat ?

Answer

To solve the problem, we need to find the speed of the boat that is moving with constant velocity while a key falls from a height of 45 meters above the water. The boat is initially 12 meters away from the point of impact when the key is released.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Height of the bridge (h) = 45 m
- Initial distance of the boat from the point of impact (d) = 12 m
- Acceleration due to gravity (g) = 9.8 m/s² (downward)
- Initial velocity of the key (u_key) = 0 m/s (since it is dropped)
- Speed of the boat (u) = ? (this is what we need to find)

2. **Calculate the Time Taken for the Key to Fall:**
The key falls under the influence of gravity. We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \(s = 45\) m, \(u = 0\), and \(a = g = 9.8\) m/s². Plugging in the values:
\[ 45 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
Simplifying this gives:
\[ 45 = 4.9 t^2 \]
Rearranging for \(t^2\):
\[ t^2 = \frac{45}{4.9} \approx 9.18 \]
Taking the square root:
\[ t \approx \sqrt{9.18} \approx 3.03 \text{ seconds} \]

3. **Calculate the Distance Covered by the Boat:**
The boat travels a distance of 12 meters in the same time \(t\):
\[ d = u \cdot t \]
Rearranging for \(u\):
\[ u = \frac{d}{t} = \frac{12}{3.03} \approx 3.96 \text{ m/s} \]

4. **Final Calculation:**
Rounding off, we can say that the speed of the boat is approximately:
\[ u \approx 4 \text{ m/s} \]

### Conclusion:
The speed of the boat is approximately **4 m/s**.

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