Question

projectiles launched horizontally

1. a baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. how fast was the ball rolling?
2. 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. when the cat slid off the table, what was its speed?
3. 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?
4. 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: Analyze vertical - motion

The vertical - motion of the objects is a free - fall motion. The initial vertical velocity $v_{0y}=0$. The vertical displacement $y$ is given by the equation $y = v_{0y}t+\frac{1}{2}gt^{2}$. Since $v_{0y} = 0$, we have $y=\frac{1}{2}gt^{2}$, and we can solve for the time $t$: $t=\sqrt{\frac{2y}{g}}$, where $g = 9.8\ m/s^{2}$.

Step2: Analyze horizontal - motion

The horizontal motion is a uniform - motion with constant velocity. The horizontal displacement $x$ is given by the equation $x = v_{0x}t$, where $v_{0x}$ is the initial horizontal velocity. We can solve for $v_{0x}$: $v_{0x}=\frac{x}{t}$.

Question 1

• Calculate the time of fall:
• Given $y = 0.70\ m$ and $g = 9.8\ m/s^{2}$, from $t=\sqrt{\frac{2y}{g}}$, we have $t=\sqrt{\frac{2\times0.70}{9.8}}\ s\approx\sqrt{\frac{1.4}{9.8}}\ s\approx0.38\ s$.
• Given $x = 0.25\ m$ and $t\approx0.38\ s$, from $v_{0x}=\frac{x}{t}$, we have $v_{0x}=\frac{0.25}{0.38}\ m/s\approx0.66\ m/s$.

Question 2

• Calculate the time of fall:
• Given $y = 1.0\ m$ and $g = 9.8\ m/s^{2}$, from $t=\sqrt{\frac{2y}{g}}$, we have $t=\sqrt{\frac{2\times1.0}{9.8}}\ s\approx\sqrt{\frac{2}{9.8}}\ s\approx0.45\ s$.
• Given $x = 2.2\ m$ and $t\approx0.45\ s$, from $v_{0x}=\frac{x}{t}$, we have $v_{0x}=\frac{2.2}{0.45}\ m/s\approx4.89\ m/s$.

Question 3

• Calculate the time of fall:
• Given $y = 5.4\ m$ and $g = 9.8\ m/s^{2}$, from $t=\sqrt{\frac{2y}{g}}$, we have $t=\sqrt{\frac{2\times5.4}{9.8}}\ s\approx\sqrt{\frac{10.8}{9.8}}\ s\approx1.05\ s$.
• Given $x = 8.0\ m$ and $t\approx1.05\ s$, from $v_{0x}=\frac{x}{t}$, we have $v_{0x}=\frac{8.0}{1.05}\ m/s\approx7.62\ m/s$.

Question 4

• First, calculate the time of fall:
• Given $y = 2.7\ m$ and $g = 9.8\ m/s^{2}$, from $t=\sqrt{\frac{2y}{g}}$, we have $t=\sqrt{\frac{2\times2.7}{9.8}}\ s\approx\sqrt{\frac{5.4}{9.8}}\ s\approx0.74\ s$.
• The speed of the pelican (and the initial horizontal speed of the fish) is $v_{0x}=7.62\ m/s$ (from question 3).
• Using $x = v_{0x}t$, we have $x=7.62\times0.74\ m\approx5.64\ m$.

Answer:

1. Approximately $0.66\ m/s$
2. Approximately $4.89\ m/s$
3. Approximately $7.62\ m/s$
4. Approximately $5.64\ m$