Question

show all work. include units.

1. the royal gorge bridge in colorado rises 321 m above the arkansas river. suppose you kick a little rock horizontally off the bridge. the rock hits the water such that the magnitude of its horizontal displacement is 45.0 m. find the speed at which the rock was kicked. ($v_x$)
2. a baseball rolls off of a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. how fast was it rolling? ($v_x$)
3. a cat chases a mouse across a 1.0 m high table. the mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. what was the cats speed when it slid off the table?($v_x$)
4. 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 pelicans initial speed? ($v_x$)
5. if the pelican in item 4 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? ($d_x$)

Explanation:

Step1: Analyze vertical - motion for time

In the vertical - direction, the initial vertical velocity \(v_{0y}=0\ m/s\), and the motion is a free - fall motion. The vertical displacement \(y\) is given by the equation \(y = v_{0y}t+\frac{1}{2}gt^{2}\). Since \(v_{0y} = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\), and we can solve for the time \(t\) as \(t=\sqrt{\frac{2y}{g}}\), where \(g = 9.8\ m/s^{2}\).

Step2: Analyze horizontal - motion for initial horizontal velocity

In the horizontal direction, there is no acceleration (\(a_x = 0\ m/s^{2}\)), and the horizontal displacement \(x\) is given by the equation \(x=v_{0x}t\). Since \(a_x = 0\), the horizontal velocity \(v_{x}\) is constant, and we can solve for \(v_{0x}\) as \(v_{0x}=\frac{x}{t}\).

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Step1: Calculate the time of fall

The vertical displacement \(y = 321\ m\). Using \(t=\sqrt{\frac{2y}{g}}\), we substitute \(y = 321\ m\) and \(g = 9.8\ m/s^{2}\).
\[t=\sqrt{\frac{2\times321}{9.8}}\approx\sqrt{\frac{642}{9.8}}\approx\sqrt{65.51}\approx 8.09\ s\]

Step2: Calculate the initial horizontal velocity

The horizontal displacement \(x = 45.0\ m\). Using \(v_{x}=\frac{x}{t}\), we substitute \(x = 45.0\ m\) and \(t\approx8.09\ s\).
\[v_{x}=\frac{45.0}{8.09}\approx5.56\ m/s\]

Step1: Calculate the time of fall

The vertical displacement \(y = 0.70\ m\). Using \(t=\sqrt{\frac{2y}{g}}\), we substitute \(y = 0.70\ m\) and \(g = 9.8\ m/s^{2}\).
\[t=\sqrt{\frac{2\times0.70}{9.8}}\approx\sqrt{\frac{1.4}{9.8}}\approx\sqrt{0.143}\approx0.38\ s\]

Step2: Calculate the initial horizontal velocity

The horizontal displacement \(x = 0.25\ m\). Using \(v_{x}=\frac{x}{t}\), we substitute \(x = 0.25\ m\) and \(t\approx0.38\ s\).
\[v_{x}=\frac{0.25}{0.38}\approx0.66\ m/s\]

Step1: Calculate the time of fall

The vertical displacement \(y = 1.0\ m\). Using \(t=\sqrt{\frac{2y}{g}}\), we substitute \(y = 1.0\ m\) and \(g = 9.8\ m/s^{2}\).
\[t=\sqrt{\frac{2\times1.0}{9.8}}\approx\sqrt{\frac{2}{9.8}}\approx\sqrt{0.204}\approx0.45\ s\]

Step2: Calculate the initial horizontal velocity

The horizontal displacement \(x = 2.2\ m\). Using \(v_{x}=\frac{x}{t}\), we substitute \(x = 2.2\ m\) and \(t\approx0.45\ s\).
\[v_{x}=\frac{2.2}{0.45}\approx4.89\ m/s\]

Answer:

\(5.56\ m/s\)

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