Compressed air is used to fire a 50 g ball vertically upward from...

Question

Compressed air is used to fire a 50 g ball vertically upward from a 1.0-m-tall tube. The air exerts an upward force of 2.0 N on the ball as long as it is in the tube. How high does the ball go above the top of the tube? Neglect air resistance.

Accepted Answer

Hi everyone. In this particular practice problem, we are asked to actually calculate the maximum distance reached by the sphere above the high age where we will have a charged sphere with a mass of 15 g. Initially at rest accelerated by the means of a vertical electric force of magnitude 6.5 Newton. And when their sphere reaches a height H which is 0.5 m, the electric force will become zero and were asked to calculate the maximum distance reach by the sphere after the height H So we will model the charge sphere as a particle like object. And we will first calculate the vertical acceleration A Y over the distance age. And using a Y, we will then find the philosophy of the sphere when it reaches the height. H. So once the electric field become zero, Newton, the sphere will still continue its motion under the influence of the gravitational acceleration. Or there will there will still be the free fall motion until it actually reaches a maximum height above age at a velocity of zero m per seconds. So the vertical acceleration is calculated using youth and second law from point 0 to the point H along the vertical axis. So from I am just going to start us off with making a rough diagram of what our system will look like. So essentially, for example, if this is the vertical access, we will have the maximum height here And this is going to be the h here which is going to be 0.5 m. This is going to be a point where this is going to equals to zero m just like. So, so we will model this so that we can calculate first the vertical acceleration A Y over this distance right here. And then from there, we will find the philosophy of the sphere when it reaches this point of H. And from there, the spirit will still continue to go upwards until it reaches a philosophy of zero because there will still be a gravitational acceleration pulling the spirit down. Okay. So from .02 h or 2.0.5, there will be two forces acting on the sphere, the electric forces particularly upward, I'm gonna draw the forces acting upon it here. So there will be a vertical force acting upwards. But at the same time, there will be a gravitational acceleration still pulling it downwards. So this is going to be our F electric and the other one is just the weight equals M multiplied by G as usual just like. So, so we want to apply Newton's second law. So F F net in the Y direction will equals to the mass multiplied by A Y or acceleration in the Y direction. And if that is going to be F electric minus W or minus M times G equals M times A Y. And as I have described previously, we want to find a Y, so I'm gonna rearrange this so that we can find A Y. So Y is going to be equals to F electric minus M times G over M Just like. So, so we know all this value from the problem statement, so we will substitute those. So AY equals f electric is 6.5 Newton. Uh the M S 15 g which is essentially 15 times 10 to the power of minus three kg. And the G is the 9.81 m per second squared. And then our M is in down below is still going to be 15 times 10 to the power of minus three kg just like so, and we will get an A Y value of 33.5 m per second squared just like. So now that we have found a Y, you want to use A Y to calculate the velocity at which the sphere reaches point H. So I'm gonna indicate fee fi zero Y as the initial philosophy of the sphere. So the initial philosophy of the sphere is zero m per second assuming it's from rest or starting from rest. And when the sphere reaches this for this point of H or point of 0.5 m, it will have the philosophy of fee one Y. So this is what we want to find, which is the fee one Y we want to use just a normal kid. A Matic equation without time to find fee one. Y. So recall that one of the cinematic equation that we can use now is 51 Y squared plus fee, fee, not fee zero Y squared equals to A Y H. So rearranging things, we have 51 Y squared equals with fee fi zero Y squared being zero. Sophie one Y squared equals two A Y H just like so and FI one Y is then going to be the square root of Two multiplied by 33.5 m/s squared multiplied by 0.5 m Just like so and that will give us the value of 31 Y of 5. m/s. That will be the velocity when the sphere actually reaches the point of 0.5 m. So once the electric field become zero, the sphere will be subjected still to gravitational acceleration. So at a point of above 0.5 m, this is going to be our system right here. There will only be one force acting upon it because the electric force is zero. So we will still have the M times the G. So in this case, we want to still use the Kinnah Matic equation without time to find the maximum distance traveled above age. So we can actually um use this same equation. But in this case fee to Y is going to be the um the velocity at the maximum point that the spirit can reach, which is going to be zero m per second. Because at the maximum point there, the velocity will will momentarily become zero. So we want to use the same equation fee to Y square minus V one Y squared equals two A Y. And this is going to be the delta Y that we want to find or the maximum distance that it can reach. So the fee to Y squared is zero. So this is going to be minus V one Y squared, which is going to be minus 5. m per second squared equals two. Um A Y delta Y and A Y here is essentially going to just be the gravitational acceleration because that is the only acceleration acting upon our system right now. So this is two G delta Y and then this will then be minus 5. m per second squared equals two, 9.81 m per second squared delta Y. So delta Y is then going to be -5. m/s squared to fight it by two multiplied by recall that the gravitational acceleration here is pointing downwards. So we want this to be minus 9.81. So this will be minus 9.81 right here just like so, and this will essentially be 1.71 m and that will be the maximum distance that the sphere can reach above each. By considering that after it reaches age, there will no electric force left acting upon our sphere. So a maximum height of 1.71 m will be reached and that will correspond to option d right here. So option D is going to be the answer to our particular practice problem and I'll be all for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and I'll be all for this video. Thank you.

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Compressed air is used to fire a 50 g ball vertically upward from a 1.0-m-tall tube. The air exerts an upward force of 2.0 N on the ball as long as it is in the tube. How high does the ball go above the top of the tube? Neglect air resistance.

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