David Troyer

2022-01-05

Lucy and her friend are working at an assembly plant making wooden toy giraffes. At the end of the line, the giraffes go horizontally off the edge of the conveyor belt and fall into a box below. If the box is 0.6 m below the level of the conveyor belt and 0.4 m away from it, what must be the horizontal velocity of giraffes as they leave the conveyor belt?

It takes t seconds the giraffes to reach the bottom of the belt:
$t=\sqrt{\frac{-2y}{g}}=\sqrt{\frac{-2\cdot \left(-0.6\right)}{9.8}}=0.35s$
The velocity of the giraffes must be:
${v}_{x}=\frac{x}{t}=\frac{0.4}{0.35}=1.1\frac{m}{s}$

enlacamig

Since the giraffes are going off horizontally, there is noinitial vertical velocity $y={v}_{oy}t+\frac{1}{2}{gt}^{2}$
so $t=\frac{\sqrt{2y}}{g}=\sqrt{2}\cdot -0.6\frac{m}{-9.8}\frac{m}{{s}^{2}}=0.35s$ in the air

Hope that helps

karton

user_27qwe

To solve the problem, we can use the principle of projectile motion. The horizontal velocity of the giraffes as they leave the conveyor belt can be determined using the following equation:
${v}_{x}=\frac{d}{t}$
where ${v}_{x}$ is the horizontal velocity, $d$ is the horizontal distance, and $t$ is the time of flight.
In this case, the horizontal distance is given as 0.4 m, and the time of flight can be determined using the equation:
$t=\sqrt{\frac{2h}{g}}$
where $h$ is the vertical distance and $g$ is the acceleration due to gravity.
Given that the vertical distance is 0.6 m, and the acceleration due to gravity is approximately $9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, we can substitute these values into the equation for time of flight.
$t=\sqrt{\frac{2·0.6\phantom{\rule{0.167em}{0ex}}\text{m}}{9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}}$
Simplifying the equation:
$t=\sqrt{\frac{0.12}{9.8}}\approx 0.109\phantom{\rule{0.167em}{0ex}}\text{s}$
Now, we can substitute the values of $d$ and $t$ into the equation for horizontal velocity:
${v}_{x}=\frac{0.4\phantom{\rule{0.167em}{0ex}}\text{m}}{0.109\phantom{\rule{0.167em}{0ex}}\text{s}}\approx 3.67\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Therefore, the horizontal velocity of the giraffes as they leave the conveyor belt is approximately $3.67\phantom{\rule{0.167em}{0ex}}\text{m/s}$.

star233

Step 1: Let's denote the horizontal velocity of the giraffes as ${v}_{x}$ and the vertical velocity as ${v}_{y}$. We are given that the vertical displacement ($\Delta y$) is 0.6 m and the horizontal displacement ($\Delta x$) is 0.4 m.
Using the equation of motion for vertical displacement, we have:
$\Delta y={v}_{y}·t+\frac{1}{2}·g·{t}^{2}$
where $t$ is the time of flight and $g$ is the acceleration due to gravity (approximately 9.8 m/s${}^{2}$). Since the giraffes fall vertically, their initial vertical velocity is zero (as they leave the conveyor belt horizontally). Thus, the equation simplifies to:
$0.6=\frac{1}{2}·9.8·{t}^{2}$
Simplifying further, we get:
$0.6=4.9·{t}^{2}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{t}^{2}=\frac{0.6}{4.9}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}t\approx 0.273\phantom{\rule{0.167em}{0ex}}\text{s}$
Step 2: Now, we can use the equation of motion for horizontal displacement:
$\Delta x={v}_{x}·t$
Substituting the given values, we have:
$0.4={v}_{x}·0.273\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{v}_{x}=\frac{0.4}{0.273}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}{v}_{x}\approx 1.466\phantom{\rule{0.167em}{0ex}}\text{m/s}$
Therefore, the horizontal velocity of the giraffes as they leave the conveyor belt is approximately 1.466 m/s.

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