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Carnegie Mellon University

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Problem 19 Medium Difficulty

A playground is on the flat roof of a city school, 6.00 $\mathrm{m}$ above the street below (Fig. $\mathrm{P} 3.19 ) .$ The vertical wall of the building is $h=7.00 \mathrm{m}$ high, to form a 1 -m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of $\theta=53.0^{\circ}$ above the horizontal at a point $d=24.0 \mathrm{m}$ from the base of the building wall. The ball takes 2.20 $\mathrm{s}$ to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

Answer

a. 18.1$\mathrm { m } / \mathrm { s }$
b. 1.09$\mathrm { m }$
c. 2.8$\mathrm { m }$

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Video Transcript

so here for party, we know that the horizontal displacement of the ball is gonna be dealt to access would be equaling the velocity in next direction. There isn't any acceleration in the extraction, so we don't have to account for that term multiplied by t. This would be equaling the initial velocity multiplied by co sign of Fada fate of being the angle of launch multiplied by tea. And so we can then say that the initial would be equaling Delta X over t co sign a theater and so this would be equaling 24.0 meters. This would be divided bye 2.20 seconds the time of flight times co sign of 50 three degrees and we find that the initial velocity is equaling 18.1 meters per second. This would be our first answer for part A for part B. Then we know that the vertical displacement of the ball at the instant it passes over the top wall. We can use Delta y equaling V. Why initial t plus 1/2 a sub y t squared. And so here we can say that then delta y is going to be equaling this would be 18 0.1 meters per second times sign of 50 three degrees multiplied by 2.20 seconds plus. And then we'll say 1/2 multiplied by negative 9.80 meters per second squared multiplied by 2.20 seconds. Quantity squared and this is equaling 8.9 meters. We know that the top of the wall is seven meters and so we can say that with the ball clears the top wall by 1.9 meters. So this would be our final answer for part B. Now for part C, we can say that. Then when the ball it's six meters above the ground, we can find essentially the time at which the ball crosses that wall and so we can say that first again. Delta y equals V. Why initial t plus 1/2 g t squared or acceleration y direction either way and so we can say that 6.0 meters will equal 18.1 meters per second, multiplied by sine of 53 degrees multiplied by t and then this would be minus 4.90 meters per second squared multiplied by T squared. And as you can see, this is a quadratic equation. You can use your cell function on your tear 84 85 or 89 in order to solve and hear T is equaling 0.497 seconds. So that's the time that it passes six meters on its way up, and then and then 2.46 seconds. That's the time that it passes, um, on its way down. And so we're picking this answer. And we know that the total we can say rather that the first the the second solution here is the time when it lands on the roof. And so we can say that delta t The change in time would be 2.46 the second the time that it lands on the roof, minus the time that it passes over that wall 2.20 seconds. This is equaling 0.26 seconds, and so after it passes the top of the wall, it still has a flight with a duration of 0.26 seconds. So how far beyond the wall is the ball landing? While we can solve by simply saying the horizontal displacement. This is equaling the velocity and X in the ex direction times Delta T And here we're using 0.26 seconds. So this would be 18 0.1 meters per second times co sign of 53 degrees multiplied by point to six seconds and this is giving us 2.8 meters. So essentially we can say that then if it lands 2.8 meters beyond the wall after clearing the wall by 1.9 meters, that is the end of the solution. Thank you for watching.

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