Questions: 2. A pelican flying along a horizontal path drops a fish from a height of 7.4 m . The fish travels 10.0 m horizontally before it hits the water below. What is the pelican's speed?

Solution Steps

First, we need to calculate the time it takes for the fish to fall from a height of 7.4 meters. We use the equation for the vertical motion under gravity:

\[ h = \frac{1}{2} g t^2 \]

where \( h = 7.4 \, \text{m} \) is the height, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( t \) is the time in seconds. Solving for \( t \), we have:

\[ 7.4 = \frac{1}{2} \times 9.81 \times t^2 \]

\[ t^2 = \frac{7.4 \times 2}{9.81} \]

\[ t^2 = 1.5087 \]

\[ t = \sqrt{1.5087} \approx 1.2283 \, \text{s} \]

The horizontal distance traveled by the fish is 10.0 meters. Since there is no horizontal acceleration, the horizontal speed \( v \) of the pelican is constant. We use the formula:

\[ d = v \times t \]

where \( d = 10.0 \, \text{m} \) is the horizontal distance and \( t = 1.2283 \, \text{s} \) is the time calculated previously. Solving for \( v \), we have:

\[ v = \frac{d}{t} = \frac{10.0}{1.2283} \approx 8.1427 \, \text{m/s} \]

Final Answer