Question

1. a pelican flying along a horizontal path drops a fish from a height of 5.4 m. the fish travels 8.0 m horizontally before it hits the water below. what is the pelican’s speed?
2. if the pelican in item 3 was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

Explanation:

Step1: Calculate the time of fall for the fish in item 3

Use the vertical - motion equation \(y = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\) (the fish is dropped vertically), \(y=- 5.4\ m\) (taking downwards as negative) and \(g=-9.8\ m/s^{2}\).

$$y=\frac{1}{2}gt^{2}\Rightarrow t=\sqrt{\frac{-2y}{g}}$$

$$t=\sqrt{\frac{-2\times(-5.4)}{9.8}}\approx1.05\ s$$

Step2: Calculate the pelican's speed in item 3

In horizontal motion, \(x = v_{0x}t\). We know \(x = 8.0\ m\) and \(t\approx1.05\ s\).

$$v_{0x}=\frac{x}{t}=\frac{8.0}{1.05}\approx7.62\ m/s$$

Step3: Calculate the time of fall for the fish in item 4

Again, using \(y = v_{0y}t+\frac{1}{2}gt^{2}\) with \(v_{0y} = 0\), \(y=-2.7\ m\) and \(g = - 9.8\ m/s^{2}\)

$$t=\sqrt{\frac{-2y}{g}}=\sqrt{\frac{-2\times(-2.7)}{9.8}}\approx0.74\ s$$

Step4: Calculate the horizontal distance in item 4

Since \(v_{0x}\approx7.62\ m/s\) (same speed as in item 3) and using \(x = v_{0x}t\)

$$x=(7.62)\times0.74\approx5.64\ m$$

Answer:

For item 3, the pelican's speed is approximately \(7.6\ m/s\). For item 4, the fish would travel approximately \(5.6\ m\) horizontally.