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A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of conta...

Question

A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of contact lime shown In the graph; What is the speed of the ball Immediatety eicr I Iaaves tne FechitF(N}60040Z00

A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of contact lime shown In the graph; What is the speed of the ball Immediatety eicr I Iaaves tne Fechit F(N} 600 40 Z00



Answers

A 115-g ball is traveling to the left with a speed of $30 \mathrm{m} / \mathrm{s}$ when it is struck by a racket. The force on the ball, directed to the right and applied over 21 ms of contact time, is shown in the graph. What is the speed of the ball immediately after it leaves the racket?

In this problem, we're going to apply the equations of two dimensional motion to an object undergoing the force of gravity. Under these specific conditions, and the conditions that we were given is that a ball is presumably hit off the ground are very close to the ground At an angle of 3° as shown here. And we have some blockade that has a height of .33 m. And were given that the ball was initially hit 12.6 m from that blockade. And the final bit of information that we were given is that the ball just makes it over the blockade uh during at the apex of its height. So as in the maximum height of the projectile uh will be uh the height of the blockade, it just passes over it at its maximum height. So we can write Why max of the projectile is equal to 3.0.330 m. Mhm. And we want to figure out the initial velocity of the ball in order for this to happen. So we first we write down the change in the height of um the projectile at any time T. Which would be one half times the acceleration in the Y direction, which is negative G. For an object undergoing force of gravity plus the initial velocity in the Y direction times time. And the change in the Exposition Exposition, there's no acceleration in the X direction because gravity only works vertically. So this will be V. Subzero, X times T. And if we draw the velocity initial velocity, yeah, we can find that the initial velocity in the X direction. Ah And if we draw the components, This angle has to be 30°, initial lost the magnitude of the initial velocity in the X direction will be the initial velocity times to co sign of three degrees. And the initial velocity in the Y direction will be the initial velocity times the sine of three degrees. So we can substitute, we could substitute those into this equation and then we just need to solve for the initial velocity. So this will be negative one half G. T squared plus v. Sub zero times to sign 3°, All Times T. And this is equal to d sub zero times the co sign of 30 degrees. There are 3° times time. And this will be true for anywhere within its trajectory. And we can apply this to the moment at which is in its apex, uh the apex of its trajectory. And at that trajectory or at that apex we know that the velocity in the Y direction is zero. And in general the why the velocity in the Y direction at any time T. Can be written as the initial velocity in the Y direction. Uh Plus the acceleration in the Y. Direction, which is negative G. Times T. So at the Apex this is equal to zero. So the time at which it takes the ball to get from its starting position to the apex which is just above the barricade, will be the subzero, why Is when they're when this is equal to zero. Uh And the time at which that occurs is his tea. And we want to solve for that T this tea will be equal to the sum zero Y over G. Just solving this equation for tea and we can plug this into uh this one of these equations, we can solve, we can plug this into uh the well the easier equation ah would be the change in the position. The change in the position is the initial velocity times the co sign of three degrees times the time it takes to get from the start to just above the apex, which is the initial velocity in the Y direction. So Visa zero times to sign of three degrees Oliver G. And we know that this is equal to what we were given. That it uh the total change in the position to get from the starting point to the Apex is 12.6 m. So therefore we have two factors in the initial velocity. Uh the factor the two months by together would be the quantity squared. And if we solve for the initial velocity, this will be the uh G. Times to change in the position Over the co sign of 3° Times a sign of 3° all under square root. And we know all these values. Now we can solve for the initial velocity. So this would be 9.81 m/s 2.6 m divided by The co sign of 3° times the sine of three degrees. And if we plug this into the calculator, We find that the initial velocity has to be so square root of 9.81 Times 12.6, divided by The coastline of three Times The sine of three will be About 48.6 m/s, and this is the velocity of the ball has to travel at in order for the the initial conditions to occur.

So we'll label the stationary ball rather well label the moving ball A. We'll label the stationary ball B and we can then say that a vector representation of their not momentum's. We can say that then this creates a sort of triangle, and this angle here would be 30 degrees. We can say this would be 60 degrees. And this is why. Because this vector here would represent the mo mentum of all a initial. This would be the momentum of ball a final. This will be the momentum of all the above be final. And so we can then say that the final momentum of Bobby equals the initial momentum of barley multiplied by sine of 60 degrees. We can say they both have the same mass. Therefore, I am the B final Equalling M V, a initial sign of 60 degrees. Of course, their masses cancel out and the final velocity of the previously stationary ball is equaling two 4.0 meters per second, multiplied by sine up 60 degrees. This is equaling 3.46 meters per second. Now for the final velocity of the initially moving ball, we can say that or the ball that hits the station able. We can say that then this would be equal to the initial momentum of the moving ball multiplied by co sign up 60 degrees. And so this would be again the final equaling via initial coastline of 60 degrees. And this is equaling for 0.0 meters per second again times coastline, 60 degrees. And we find that then the final velocity of ball a 2.0 meters per second again in the opposite direction. All right, and rather in the If you were to apply a Cartesian access, this was essentially being the third quadrant. That is the end of the solution. Thank you for watching.

So here, um, looking at figure 9 16 We know that the impulse Elsa F. Delta T or the forced Times the time over which the forces exerted would be equal to the area under the curve or under the graft line. And then this would be equal to the mass times the change in the velocity due to the impulse women some serum. So we can say that this would be 1/2 multiplied by 2.0 Newton's multiplied by 2.0 seconds. Of course, this is simply the area of a triangle. This would be equal to the mass times, the final velocity minus the initial velocity. And we can then say that the this would be equal toe 2.0 Newton seconds. This would be divided by the mass of 0.150 kilograms, plus the initial velocity of 12 meters per second, equaling then the final velocity. And we have a final velocity of 25 meters per second. So this would be our final answer. That is the end of the solution. Thank you for watching

Hello. In this problem we have a tennis ball player that is standing at 12.6 m from the net. So the player is standing here And he is 12.6 m away from the net. He wrecks the ball with his wreck at an angle of free degrees so that the ball moves and it barely crosses that it barely crosses the net at its epics. And the epics here means the maximum height that a trajectory can reach. Yeah. Also it is mentioned in the problem that the ball needs at least a point 33 m to cross the net. So that's the height. That's the maximum height. So H Max is equal to open 3 3 And also we have failed to that is equal to 3° and we have also The X. Which is the distance from the player to the net, which is 12.6. Although we will not use this piece of information in getting our answer. The question is to calculate the initial velocity that the ball had when it left the racket of the player. So looking at the equation of maximum height, we know that the maximum height can be getting from the I. Squared sine squared theta Divided by two G. So here we know the height already. And of course we know that gravitational exploration. We know theta. So it's straight forward to get the velocity from here. So let's rearrange this equation a little bit rearranging this. We have the I. Squared is equal to H. Multiplied by two G. Divided by sine squared theta. So blood in this on our calculator. The answer is 2361.4 and getting the square root to this. So the initial velocity Turns out to be 48.6 meters per second approximately.


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