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A cannonball is launched, and hits a target at a higherelevation than the cannon. Which of the following is correct?-The speed of the cannonball when it hits the ta...

Question

A cannonball is launched, and hits a target at a higherelevation than the cannon. Which of the following is correct?-The speed of the cannonball when it hits the target is greaterthan when it was launched.-The speed of the cannonball when it hits the target is the sameas when it was launched.-The speed of the cannonball when it hits the target is lessthan when it was launched.

A cannonball is launched, and hits a target at a higher elevation than the cannon. Which of the following is correct? -The speed of the cannonball when it hits the target is greater than when it was launched. -The speed of the cannonball when it hits the target is the same as when it was launched. -The speed of the cannonball when it hits the target is less than when it was launched.



Answers

$A$ cannon is placed at the bottom of a cliff $61.5 \mathrm{m}$ high. If the cannon is fired straight upward, the cannonball just reaches the top of the cliff. (a) What is the initial speed of the cannonball? (b) Suppose a second cannon is placed at the top of the cliff. This cannon is fired horizontally, giving its cannonballs the same initial speed found in part (a). Show that the range of this cannon is the same as the maximum range of the cannon at the base of the cliff. (Assume the ground at the base of the cliff is level, though the result is valid even if the ground is not level.)

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

In this problem. Two cannons are being shot from a 50 meter cliff, one is being shot at an ankle of 60 degrees at 80 meters per second and one is being shot horizontally at 40 meters per second. And we want to know which one has a greater displacement. Well, if we look at just the angled velocity, it tells us that we have a velocity in the ex direction of 40 meters per second, so it will have the same displacement in the ex direction as the horizontally lunched cannon. So we just need to look for our displacement in the UAE direction. So our displacement in the wind direction can be calculated using Arcana Matic equations, because we're only looking the Y direction we need to make sure we're all amusing wide values. So our initial velocity and the Y direction for the horizontally launched project hell would be zero, and so our displacement is 1/2 negative. 10 after two seconds uses a displacement of 20 meters, so we're now 20 meters lower than where we started, and then same equation. But now we're going to use the displacement, the velocity of the wind direction, which is 80 sign of data, which is 60 degrees. So we're gonna say Delta y equals my initial velocity again. We're only concerned about the Y direction. So all of our all of these values have to be in the UAE direction. So our initial velocity in the UAE direction is 80 sign of 16 times two plus 1/2 negative 10 to square gives us a displacement of 118.6 meters. So we can see that this displacement is going to be much, much greater than our displacement in the UAE direction. So we have a greater displacement by the angled Kenan, which is captain beat.

In this problem, we have to find which of the following is true. Statement regarding the motion of projectiles. Suppose this is object. It has initial speed V. It is the angle of the horizontal and it will go like this. Yeah. So the statement first says that the all projectiles have zero velocity and the effects of trajectory. We know that FX. Is that a decree? At this point, this is the maximum high point two. We know that during projectile, the horizontal component of velocity means equal during the projectile. This is equal to the cost theater. But at the apex point, Or we can say at the maximum high point, there will be no vertical component of philosophy. So the statement is wrong because it has velocity which is equal to the cost center. The statement? Yes. Is it all the statement B says the velocity of projectile is the smallest and the effects of trajectory right? It is right, because at the apex there will be no component of velocity in the vertical direction. So the statement we is right. The statement says. This says the acceleration of projector is greatest at the effects of that exactly. You know that at all the time acceleration is equal to G, which is due to gravitation. So we can say that exhalation remains same duty, projectile motion. So this is I often see is also wrong and statement These projectile have maximum kinetic energy at the apex of the trajectory. No, it has a minimum kinetic energy and the effects of that as a free so option B will be also wrong. So statement we are option. We will be our characters that thank you.

For the projectile for a projectile trajectory in the epics the lost he's momentarily zero s o. The answer is okay. Where is the acceleration is a constant. Uh, So, um, but see is wrong. Projectile maximum kind. Take energy. That packs is wrong. And but B is wrong. So what is the right answer?


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