To determine the minimum horizontal velocity required for the cliff divers to miss the rock, we first need to calculate the time it takes for them to fall from the cliff to the water below.
Given:
- Height of the cliff \( h = 34.8 , \text{m} \)
- Acceleration due to gravity \( g = 9.8 , \text{m/s}^2 \)
Using the second equation of motion, we can determine the time \( t \) it takes to fall this distance:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{g} \]
Substituting in the given values:
\[ t^2 = \frac{2 \times 34.8}{9.8} \]
Calculating it step by step:
\[ t^2 = \frac{69.6}{9.8} \approx 7.1 \]
\[ t \approx \sqrt{7.1} \approx 2.66 , \text{s} \]
Now that we have the time it takes to fall to the sea, we need to find the minimum horizontal velocity \( v \) that would allow the divers to clear the rock. The rock is a horizontal distance of \( d = 11.37 , \text{m} \) away from the base of the cliff.
The horizontal distance covered can be expressed as:
\[ d = v \cdot t \]
To find the minimum horizontal velocity, rearranging gives:
\[ v = \frac{d}{t} \]
Substituting the values we have:
\[ v = \frac{11.37 , \text{m}}{2.66 , \text{s}} \approx 4.28 , \text{m/s} \]
Therefore, the minimum horizontal velocity must be approximately:
\[ \boxed{4.28} , \text{m/s} \]
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