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Fant AAi-pound bel 8nd & {OO-pound ball are dropped from heighl 0f 10 faet at Ihe same tme. In the absonco 0f air rosistance,tha two balls end in 4 teIha iO0-po...

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Fant AAi-pound bel 8nd & {OO-pound ball are dropped from heighl 0f 10 faet at Ihe same tme. In the absonco 0f air rosistance,tha two balls end in 4 teIha iO0-pound ball hils thc ground litst Iho 1 pound ball hlts thc ground first Thoro # nol cnough intormalion dctormino whch bull hita Iho Qrourd IuelSubmllDeaueel AnawotProvide Fcedback

Fant A Ai-pound bel 8nd & {OO-pound ball are dropped from heighl 0f 10 faet at Ihe same tme. In the absonco 0f air rosistance, tha two balls end in 4 te Iha iO0-pound ball hils thc ground litst Iho 1 pound ball hlts thc ground first Thoro # nol cnough intormalion dctormino whch bull hita Iho Qrourd Iuel Submll Deaueel Anawot Provide Fcedback



Answers

From the top of a tall building, you drop two tabletennis balls, one filled with air and the other with water. Both experience air resistance as they fall. Which ball reaches terminal velocity first? Do both hit the ground at the same time?

When solving for a freefall problem, we need to look at the three different motion equations that are presented to us, you could say our final velocity is equal to our initial velocity puts acceleration times time. Our final position is equal to initial position plus initial velocity times time plus 1/2 acceleration times time squared. Or our final velocity squared is equal to our initial velocity squared, puts two times the acceleration times the change in position. Note that in none of these equations, mass is a factor. So in this circumstance where we're talking about two objects of one kg ball and a 10 kg ball, both being dropped simultaneously from the top of um some position and dropped the same height when asked which will hit the ground first. We would first of all use This # two equation. Um the object is being dropped so its initial instantaneous velocity is going to be zero. So we would say that in both cases both objects have the same initial position. They have the same acceleration and following for the same time. So therefore see both will land at the same time.

So we're told we have a happy and a sad ball and we dropped him from the same height. Um, one ball hits the ground and rebound almost to the original height. The other ball does not bounce on, represent each process with the bar chart starting just before the balls hit the ground just after the ball 2nd 1st bar rebounds and the second ball stops and choose the ball as the system. So here we have our two balls just before they hit the ground, moving at the same velocity, presumably, and I'm assuming that the happy ball rebound and the sad balls blots on the ground doesn't rebound. So if we if we look at the impulse momentum, so initially, thing have a d a initial momentum. And again if we say positive is upwards and I'm going using a plus sign for the happy ball, any minor aside for sad ball. So if the momentum is initially down and we said positive was up, then we have initial negative initial momentum. We have some momentum from the ground, um, acting so, uh, we have when it hits this, we have a force from the ground acting that causes an impulse from the ground. And then we also have the impulse from gravity that's acting too. Keep the ball moving downwards. But if way fit, rebounds, then then it reaches the same height than the velocity at which it rebounded should be the same as their last, which would hit. So this should be the just that simply the equal and opposite to this. And so that means this this impulse need to overcome this momentum, and this involves to create this momentum and for the sad ball we have again, its initial mo mentum is the same as other case, assuming they weigh the same, obviously. Then again, we have a an impulse from the ground when it hits. And then we have the impulse from gravity over the time, same time period, and because it doesn't rebound, and it just comes to rest. We know that this impulse needs to offset this momentum and this impulse. So we have. You know, in this case, this is a big zero here, and so this and this this needs to offset this in this to create that zero here

Okay, So in this question, we're looking at which situation has the greatest in falls. Let's begin by reminding ourselves with definition of impulse, and that is that F Delta T force Times Time is equal to M. Delta Bi. So that is the impulse is equal to the change in momentum. Now we've been given in the question that both the balls have the same mass. So you know that I m is the same for both situations. That means when we're looking at, which has the greatest impulse, the thing we really interested in is which has the greatest change in velocity. So it's just labeled these situations a and B for ease on. Now let's look at the change in the musty so in the situation. A. It's accelerating as it's falling to reaching the final Boston. And then it's stopping. So the initial blasting is the velocity just before it collides with the four on, let's call that the it is then becoming stationary. It is going from the it is there now. When we do change in velocity, we do the final minus the initial, so that means we've got zero minus V and that gives us minus V, so the change in velocity for a is minus would be. Now let's look at situation. The situation be is doing the same. It's falling, it's accelerating. It's reaching of lost e just before it hits the ground. And that is gonna be the same velocity. So it's going from the then hitting the ground and bouncing back up so that it's coming back up with a velocity. So because it's going back up, we've got to change the sign in our velocity on. We don't know the speed, its coming back up. So let's just call it letter you, for example. So we're gonna do the same thing again. The change in velocity is the final minus the initial. So that is minus you minus V, which equals minus B plus you. So when we look at the magnitudes of these two, change in velocities, we can see that the magnitude of the change for B is clearly greater than it is for a excuse. We've got this extra additional turn you so because we know that the change in velocity is bigger for situation be, then the change in momentum is going to be bigger for situation be. And as we know, that change in momentum is impulsive. Then the bowl, no obey experiences, a greater impulse.

A juggler tossing a set of balls upward, and we're told he could toss a maximum of 10 balls up and that it would take 2.5 seconds in order to do the ball would have to be in the air in order for him to do that. And that really means 1.25 seconds, up 1.25 seconds down. The height of any ball at any time is negative. 16 T Square plus V subzero times the time the speed of any ball is the derivative of that negative 32 t plus v subzero. We do know that at a time of 1.25 seconds, the speed is zero. So let's use this equation will say that the speed of zero when t is 1.25 Well, negative 32 times 1.25 is negative 40. So that lets us know that the initial velocity after we had the 40 was 40 feet per second. So that's the speed that he would have to toss it at now. How high would the ball go at that speed? Well, let's go to wife T again. This is for a time of 1.25 seconds. So I'll fill in 1.25 seconds for the time and 40 seconds for V subzero. Successfully filling that in gives me a height of 25 feet. So he's gonna have to be able to talk to toss those balls up 25 feet. Well, what happens if you were thrown 11th ball? An 11th ball is gonna add on additional half Second, because again, a catching the tosses 1/2 2nd So now the balls in the air for three seconds, which means 1.5 seconds up and 1.5 seconds down. So we're going to be using the same equations we have before. Just different The sub zeros position function negative. 16 t squared, plus visa bough t velocity function derivative of that negative 32 t plus V subzero. After 1.5 seconds, the velocity is zero. So if let why prime equals zero when t is 1.5 zero equals negative 48 plus a V subzero and adding the 48 gives us an initial velocity of 48 so we can see that the ball at an increase of speed from 40 to 48. Well, how much higher is this gonna make the ball go? Lau. Let's look at the position function again. We're doing this for a time of one 0.5 seconds, so that's going to be negative. 16 times, 1.5 squared, plus 48 times 1.5, which will give us Ah, height of 36 feet, which is 11 feet higher than we had for the 10 ball toss.


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