Question

birdman is flying horizontally at a speed of 28 m/s and a height of 64 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? scoreboard horizontal distance (dx) horizontal velocity (vx) height: 0 m check answer

Explanation:

Step1: Find the time it takes for the turd to fall

The vertical - motion of the turd is a free - fall motion. The height \(h = 64\ m\), and the equation for vertical displacement in free - fall is \(h=v_{0y}t+\frac{1}{2}gt^{2}\). Since the initial vertical velocity \(v_{0y} = 0\ m/s\), the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(g = 9.8\ m/s^{2}\). Solving for \(t\), we get \(t=\sqrt{\frac{2h}{g}}\).
\[t=\sqrt{\frac{2\times64}{9.8}}\]

Step2: Find the horizontal distance

The horizontal motion is a uniform - motion with a constant horizontal velocity \(v_x=28\ m/s\). The horizontal distance \(d_x\) is given by the formula \(d_x = v_x\times t\). Substitute \(t=\sqrt{\frac{2\times64}{9.8}}\) into the formula:
\[d_x=28\times\sqrt{\frac{2\times64}{9.8}}\]
\[d_x = 28\times\sqrt{\frac{128}{9.8}}\approx28\times3.62\approx101.36\ m\]

Answer:

\(101\ m\) (rounded to the nearest whole number)