Question

1. a pelican flying along a horizontal path drops a fish from a height of 25.4 m. the fish travels 18.0 horizontally before it hits the water below. what was the pelicans speed (vx)?
2. what was the speed of the fish as it hit the water?

vx =
vy =

Explanation:

Step1: Calculate time of fall

\(t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2\times25.4}{9.8}}\approx2.28\ s\)

Step2: Calculate horizontal speed

\(v_x=\frac{x}{t}=\frac{18.0}{2.28}\approx7.89\ m/s\)

Step3: Calculate vertical speed at impact

\(v_y = gt=9.8\times2.28 = 22.34\ m/s\)

Step4: Calculate total speed at impact

\(v=\sqrt{v_x^{2}+v_y^{2}}=\sqrt{(7.89)^{2}+(22.34)^{2}}\approx23.7\ m/s\)

Answer:

1. First, find the time it takes for the fish to fall vertically:

• The vertical - motion of the fish is a free - fall motion. The vertical displacement \(y = 25.4\ m\), and the initial vertical velocity \(v_{0y}=0\ m/s\). The equation for vertical displacement in free - fall is \(y = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\).
• Solving for \(t\), we get \(t=\sqrt{\frac{2y}{g}}\), where \(g = 9.8\ m/s^{2}\).
• Substitute \(y = 25.4\ m\) into the formula: \(t=\sqrt{\frac{2\times25.4}{9.8}}\approx\sqrt{\frac{50.8}{9.8}}\approx\sqrt{5.184}\approx2.28\ s\).

1. Then, find the horizontal speed of the pelican (which is the same as the initial horizontal speed of the fish):

• The horizontal displacement \(x = 18.0\ m\). In horizontal motion (assuming no air - resistance), there is no acceleration (\(a_x = 0\ m/s^{2}\)), and the horizontal speed \(v_x\) is constant. The equation for horizontal motion is \(x = v_x t\).
• Solving for \(v_x\), we get \(v_x=\frac{x}{t}\).
• Substitute \(x = 18.0\ m\) and \(t\approx2.28\ s\) into the formula: \(v_x=\frac{18.0}{2.28}\approx7.89\ m/s\).

1. Next, find the vertical speed of the fish as it hits the water:

• Use the equation \(v_y=v_{0y}+gt\). Since \(v_{0y} = 0\ m/s\), \(v_y=gt\).
• Substitute \(g = 9.8\ m/s^{2}\) and \(t\approx2.28\ s\) into the formula: \(v_y=9.8\times2.28 = 22.34\ m/s\).

1. Finally, find the speed of the fish as it hits the water:

• The speed \(v\) of the fish as it hits the water is given by the Pythagorean theorem \(v=\sqrt{v_x^{2}+v_y^{2}}\).
• \(v_x\approx7.89\ m/s\) and \(v_y = 22.34\ m/s\).
• \(v=\sqrt{(7.89)^{2}+(22.34)^{2}}=\sqrt{62.25 + 499.08}=\sqrt{561.33}\approx23.7\ m/s\).

The pelican's speed \(v_x\approx7.89\ m/s\), the vertical speed of the fish as it hits the water \(v_y = 22.34\ m/s\), and the speed of the fish as it hits the water \(v\approx23.7\ m/s\).