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A daredevil is shot out of a cannon at $45.0^{\circ}$ to the horizontal

with an initial speed of $25.0 \mathrm{~m} / \mathrm{s}$. A net is positioned a horizontal distance of $50.0 \mathrm{~m}$ from the cannon. At what height above the cannon should the net be placed in order to catch the claredevil?

$h = 10.8 \mathrm { m }$

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Cornell University

University of Washington

Hope College

McMaster University

So the initial components for the deal there devil we can save the ex initial is gonna be equaling the why initial in this in this specific case. Because this is get simply gonna be equaling v initial co sign of 45 degrees or the initial sign of 45 degrees. Knowing that, of course, sign of 45 degrees is equaling to co sign of 45 degrees. This will essentially be equaling 25.0 meters per second, divided by radical too. And so we can say the time to the time required to travel 50 meters horizontally would be Delta X divided by the ex initial. And this is equaling 50 0.0 Mears multiplied by radical too divided by 25.0 meters per second. And this is giving us too radical. Two seconds. The vertical displacement now of the dare of devil at this time is gonna be Delta. Why, this would be equaling V Y initial t plus 1/2 times the acceleration in the UAE direction Times t squared. And so this is gonna be equaling 25.0 meters per second. Divided by radical too. We're gonna multiply this by too radical, too. Seconds minus 1/2 multiplied by 9.80 meters per second squared multiplied by two radical two seconds. Quantity squared. And so we find that then delta Y is equaling 10.8 meters. This would be the proper heights above the level of the cannon to place the net 10.8 meters. This would be our final answer. That is the end of the solution. Thank you for watching.