Answer:
19.99 m/s when it hits the rocks below.
Explanation:
First, we need to find the time it takes for the clam to fall from a height of 25 meters. We can use the equation:
\[ h = \frac{1}{2} g t^2 \]
Where \( h \) is the height, \( g \) is the acceleration due to gravity (approximately 9.8 m/s^2), and \( t \) is the time.
Plugging in the values, we have:
\[ 25 = \frac{1}{2} \times 9.8 \times t^2 \]
Simplifying the equation, we get:
\[ t^2 = \frac{25 \times 2}{9.8} \]
\[ t^2 = \frac{50}{9.8} \]
\[ t \approx 2.04 \, \text{s} \]
Now, we can use the time and the height to find the velocity of the clam when it hits the rocks. We can use the equation:
\[ v = g t \]
Plugging in the values, we have:
\[ v = 9.8 \times 2.04 \]
\[ v \approx 19.99 \, \text{m/s} \]
Therefore, the clam is going approximately 19.99 m/s when it hits the rocks below.