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Two cannons are fired from a cliff at a height of 50 m from the ground. Cannonball A is fired horizontally with an initial velocity of 40 m/s. Cannonball B is fired at a launch angle of $60^{\circ}$ with an initial velocity of 80 m/s. Which cannonball will have a greater magnitude of displacement after two seconds?(A) Cannonball A(B) Cannonball B(C) Both cannonballs will have the same displacement.(D) Cannot be determined

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Chapter 13

Practice Test 3

Section 1

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hi. In the given problem, two cannonballs ah projected from the same height and the height from which they were projected. He's given us 50 meter now one off the cannonball, which is leveled as a is thrown horizontally. Then, on initial speech off you A is equal to 14 m per second and another can involve is projected. You can initially speed UV off 80 meter per second at an angle off 60 degree from the horizontal. No, after a given time interval off 2.0 2nd, we have to find which off the cannonball will be having the more displacement displacement from the initial position. So to find it. First of all, we consider Cannonball he for this cannonball as it is moving horizontally only so it's distance traveled horizontally will be found simply using distance. He goes to speed into time and the speed waas 40 m per second at the time given here is 2.0 seconds, so this distance comes out to be empty meter now, for the distance traveled by it vertically means why it for it. We will use second equation of motion which says y a is equal to you. a y means initial velocity of cannon ball a in vertical direction into T plus half GT Square as there was no initial component vertical component. So this is zero less half into 10 for acceleration due to gravity. Again square off to canceling this too. We get 20 meter so it's net. Displacement here will come out to B R. A. Is equal to x a square, plus why a square or we can say this is square off 80 plus square off 20 or this is 6400 plus 400 which comes out to be 6800 m. No for another can involve cannonball bait. Its initial horizontal velocity will be given by you be intercourse 60 degree means 80 m per second into one by two for cost 60 degree. So it comes out to be 40 m per second. So horizontal distance traveled by this cannonball also will be equal to will be given by you be X means to speed into time. So this is 14 m per second time again two seconds. So this is also 80 m. But as far as its vertical distance traveled is concerned that will be given by the second equation of motion you b y t plus half g t square. So it comes out to be 80 meter per second into sign 60 degrees into time T minus half into 10 the square off to We have used negative sign as the ball is initially moving in upward direction, for which the gravity will be taken as negative. Now here it becomes 80 into room three by two for science, 60 degrees into minus 20. Nita. So here it becomes 80 Route three minus 20 meter, which comes out to be 118 0.56 Neater so it's displacement from initial position will be given by that's quite off X B and the squared off y b the tradition and then the square root. So as XB, he's having the same value. 80. But why be? Is having much more value than that off by a Hence here, we can say without calculating it, we can conclude that R B is more than are a means. The distance, the total displacement net displacement for cannonball be is more than that. Hey, and this is the answer for this given problems. So here we can see option the his correct Thank you

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