Question

birdman is flying horizontally at a speed of 20 m/s and a height of 55 m. birdman releases a turd directly above the start of the field. how far from the start of the field should the robot hold the bucket to catch the turd?

Explanation:

Step1: Calculate time of fall

The vertical - motion is a free - fall motion. The height \(h = 55m\), and the equation for vertical displacement in free - fall is \(h=v_{0y}t+\frac{1}{2}gt^{2}\). Since the initial vertical velocity \(v_{0y} = 0m/s\), the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(g = 9.8m/s^{2}\). Solving for \(t\), we get \(t=\sqrt{\frac{2h}{g}}\).
\[t=\sqrt{\frac{2\times55}{9.8}}\]

Step2: Calculate horizontal distance

The horizontal motion is a uniform - motion with constant velocity \(v_x=29m/s\). The horizontal distance \(x = v_x t\). Substitute \(t=\sqrt{\frac{2\times55}{9.8}}\) into the equation for \(x\).
\[x = 29\times\sqrt{\frac{2\times55}{9.8}}\]
\[x=29\times\sqrt{\frac{110}{9.8}}\approx29\times\sqrt{11.22}\approx29\times3.35\approx97.15m\]

Answer:

\(97.15m\)