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Carnegie Mellon University



Problem 56 Hard Difficulty

Antlion larvae lie in wait for prey at the bottom of a conical pit about 5.0 $\mathrm{cm}$ deep and 3.8 $\mathrm{cm}$ in radius. When a small insect ventures into the pit, it slides to the bottom and is seized by the antlion. If the prey attempts to escape, the antlion rapidly launches grains of sand at the prey, cither knocking it down or causing a small avalanche that returns the prey to the bottom of the pit. Suppose an antlion launches grains of sand at an angle of $72^{\circ}$ above the horizon. Find the launch speed $v_{0}$ required to hit a target at the top of the pit, 5.0 $\mathrm{cm}$ above and 3.8 $\mathrm{cm}$ to the right of the antlion.


$t _ { 0 } = 1.04 \mathrm { ms } ^ { - 1 }$


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Video Transcript

So we're gonna use the exposition to solve for time first or the X the kinetics in the extraction and then Killem attics in the wind direction. So we can first say that Delta axes equaling V X initial times t. And we can say that then there isn't any acceleration in the, um, ex direction, so we don't have to account for that term. And we can simply say that t is gonna be equaling Delta X over the initial co sign of data. And so Ty's gonna be equaling 3.8 centimeters divided by the initial we can save be not co sign of seven 72 degrees. And so we can say that that Delta why is equaling uh, the why initial t plus 1/2 GT squared. Here we're going. Teoh say that upwards is positive. So we can say that then five centimeters is gonna be equaling the not sign of 72 degrees multiplied by t of 3.8 centimeters, divided by the initial coastline of 72 degrees. And then this would be plus 1/2 most applied by negative 980 centimeters per second squared so well, we'll just convert everything. Two centimeters, um, multiplied by 3.8 centimeters, divided by the initial co sign of 72 degrees quantity squared. And so solving For the initial, we can use your tea I 84 85 or 89 in order to solve for the initial. And we find that the initial is equaling 105 centimeters per second. This would be the launch speed. This should be our final answer. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
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