Ball hits the water in 5.2s, calculate the horizontal distance traveled by the cannon ball. So. That's what they want from a the horizontal distance, the height of the Cliff. B and the angle has which the cannon ball enters the water. Okay. So how would you tackle this? The horizontal distance traveled by the carnonball. Horizontal distance, 3m per of that is 30m per second of a Cliff and 5.2s horizontal distance. So we should know about the time time Yeah, 5.2s. And we should also want to we also want to know about the horizontal velocity so we can use a triangle on no, we can't use we can't use triangle. 好，okay. 30 times 5.2. 156. 156m horizontal distance. Good, the height of the Cliff height. High of the Cliff, okay, the final velocity is zero, the t is 5.2s and the acceleration is 9.81. And we want to know the this, want to know the vertical distance and in the height. So. S equal to V T minus a half 80. So that is a half times 9.81 times 5.2. 25.5. It's you square t. No. So 5.2 squared times 9.81 times 0.5. 133 133m good. The angle with which the cannon ball enters the water. So this angle. So we know the horizontal velocity as 30m per second. We can work out the vertical velocity by using the. Equals U. S. At. So U is zero so it's 9.81. Times 5.2. 9.81 times 5.2 so. So that's 51m per second. So if we know two velohies, we can find this angle. Or this angle. Okay, so if we say ten theta. Is 51 over 30 sheet. So fasia is inverse ten, so 51 divided by 30. In verse ten, so 59.5. Remember, velocity is a vector quantity, so we can use Pythagoras, we can use our trigonometry to find the angle, okay? Okay, Jackson. Yeah. Rise. Now if you remember, we were looking at this question and we had a non projectile motion, which means its acceleration is not uniform. So we've answered all of these questions except this bit. The photograph shows a shutlecock. The shuttle always moves through the air with the feathers trailing behind. So the direction will be like this. The heavy bit will be nose. First, explain how the feathers affect the motion of the shuttlecock along its path for three Marks x. What could we say here? So explain means why is something happening? And how? Why? How? Okay, you're asked to explain. Describe means what, when, where. Whereas explain means why does it happen and how does it happen? Why and how? Affect the the feathers affect the motion of the shutle cook along its path. This is your shutcoshutle cookay. Yeah, I see. Explain hunting. When the shudow clock ck我shadow Yeah shudow cock alone is path，there's an air resistance and make. Make the make the friction with the feather, so the feather will make thermal energy. And is that a will? And that. And that is the way you transferred by the kinetic energy. There is. The pair the shuttle fuck. Connection energy to the air. Yeah. Others will call turbulence. Remember what turbulence, the opposite of laminar flow. So laminar flows in layers, turbulent flow is when it's erratic. This results in drag forces, horizontal velocities, y's decrease. So acceleration is no longer uniform. I'm sure something like that. What you've puwould get some Marks. So acceleration is no longer uniform. But certainly the heating, the air drag, hair resistance. We mentioned, so you get at least part of that question. Okay, here is another PaaS paper question. Jackson, again, it involves, it's a bit like our cannonball when we started with in a castle overlooking a river, a cannon was once employed too far at enemy ships. One ship was hit by a cannon ball at a horizontal distance of 150 mefrom. The cannon, as shown in the figure, the height of the cannon above the river was 67m, and the cannon ball was fired horizontally. So we know that distance. That distance show that the time taken for the cannon ball to reach the water surface after being fired from the cannon was about 3.7s. The so the time taken. I'm taken. Okay. So the time taken that is. There were no distances. S is equal to U, T, plus a half A, T squared. We know a, we know U is zero, we know s, so we could properly use this equation. Using the height from the river. So. So s equals U T. Plus a half at squared. So we know. The vertical drop is 67m. We know U is zero. And we know A A half A T squared and that's 9.81. Okay, so. Do we get 3.7s for tea? Can we check that? So 6767 times two diviided by 9.81. 11.7 square root ten, no 13, 13.7, 13.7. Okay, so it's t squared, so square root that. 3.7, 3.7. Okay? So when you're asked to show that, you must show your working, okay? So don't doin your head. You must, when you're asked to show something, the examiner will expect to see that. Calculate the velocity at which the cannon ball was fired. Give your answer to an appropriate number of significant figures. So. What was the initial velocity? So we know t now can we use this v is equal to U plus at t we know a, we know t. So e squared is equal to U squared plus two as we know s. Be his distance over time. Okay, so they've just used v equals s over t. Yeah. So v, they've effectively used that equation. So they've taken the horizontal distance divided by the time, which is 3.7. The time we've just worked out. So 150 divided by 3.7 is 44.5. 40 15e. Tuper, second. 41m per second. Because this is to one decimal place. That's why I gave that to one decimal place. Calculate the vertical component of velocity just before the cannon ball hits the ship. So what is this? Philosopcity. Just before it hits the ship. Vertical component of velocity, component of velocity just before the can. The. So maybe that is the final velocity. The final nel Veloso, the time is 3.7, and the vertical velocity is that vertical, we vertical s equal to 67n equal to 9.81, 67, equal to V, T V, and 3.7 minus a half times, 9.81 times t square, 3.7 square. V equal to 36.3m per second. I'll just check the answer. Yeah. By calculation or scale drawing, find the magnitude and direction of the velocity of the cannonball just before it hits the ship. Graitude and direction of the velocity of the cannonball just before it hits the. So we know the vertical component. We know the. Horizontal component. So we can find the resultant. And an angle. So. This is 36. This is 41. So to get this resultant, we use Pythagoras and then we use trick to resolve it. So square root of 41 squared plus 36 squared. So that's the resultant. Yeah the root 41 square plus 36 square 4054.6Yeah. And the direction. Direction we can use tangent. 36 divided by 41. 41.3. 54 okay, calculate the loss and gravitational potential energy of the cannonball mat if the MaaS is that. What's the equation for gravitational potential energy? Potential energy m. So in the whole process, it is gravitational potential energy transfer to the kinekinetic energy. So 22. Kiyeah kilograms. 22 times 9.81 times 67. 1.4 times ten years of four. Describe the energy changes. So describe so this is waswhen where what are the energy changes? When do they occur? And where? So what energy changes take place, the cannon ball, as it takes this path? First of all, what does the gravitational potential energy get transferred to? Most of the gravitational potential energy will be transferred to what Jackson. Netic energy Yeah. But yes, that is two Marks. So maybe Yeah nicso mostly the gravitational potential energy will transfer to the connecting energy, but air resistance will make friction with the canon bow and make thermal energy. So that will lead to the loss of the gravitational potential energy. Come mark. Gravitation potentially Yeah internal heat energy. So they were just looking for the heat energy. That is an no cake question. Okay, so a bullet is fired horizontally at 200m per second, at a target 40m away, neglecting air resistance. How far below the point of aim will it hit the target? So we have a bullet. Its initial velocity is 200m per second. And it travels to a target which is 40m away. So target here. So neglecting air resistance. So that means we can ignore air resistance. How far below the point of aim will it hit the target? So if it kept straight, it would hit the target there that it drops somewhat. So ittake a slightly parabolic path and drop. So time is distance over speed. So we can work out the time it's in the air. During this time, the bullet has been accelerated downwards. So it's vertical velocity after time t, so v is U plus at. So initial velocity is that and a we know. So. The vertical component is independent of the horizontal component, so we can work out how far it drops. How far it drops because it's accelerating downwards. S is equal to U, T plus one half at squared. So the vertical motion, initial vertical velocity, so s is equal to I, half a two squared, which is 0.196m. So it falls 19.6 cm as it travels 40m distance away. 0.196. So I suppose when you're firing a pistol, youhave to aim slightly above it. If you wanted to hit a bullseat, the center of the target. Okay. To find the angle of the horizontal. So we know the vertical drop, the vertical speed, the horizontal speed. So again, using ten, we can get that, okay? This is question twelve. Yeah, so a skydiver made a skdive from an aeroplane e the graph shows how the velocity v of the skydiver varied with time, from the instant she left the plane to the instant just before the parachute opened. So sheopen her parachute there. Determine the acceleration of the skydiver when t is 4s. Acceleration. So this is v, this is t, so what's the correct way to find the acceleration, the instantaneous v divided by two? Yeah but it would be better to take a tangent from that point. Oh Yeah, Yeah, so. Two, six, so four. And this is. 22 and a half to 30 so 7.5. Okay, I mean, so delta y is 7.5 divided by four. So 1.875. So for three Marks, if you show this on the graph, youget the Marks. That's really what they're wanting you to see because. They want you to recognize that it's tangent drawn at 4s. 1.5. 1.4 to 1.6. Well, we didn't quite we didn't make our tangent big enough. So they want you to do probably this. To get this the answer. Okay, okay, so draw this draws this huge triangle. Yeah the bigger the triangle, the better, the more accurate the answer. See when I did it, I got out of the range of accuracy. They're expecting an answer from 1.4 to 1.6. So if I use my second tangent, let me see what we get. So up to 8.5. 8.622 and this is 15. So 15 divided by 8.6 and I still get. If I made it to nine. 15 divided by nine is 1.66. So that's better. 15 divided by 91.7. Yeah. So they're actually giving accuracy Marks there. Determine an approximate value of the displacement of the skydiver over the first 16s of the skydive. Do you remember how we worked out the displacement from a velocity time graph, Jackson? Veltime graph. Yeah I so we have velocity against time. Yeah how would we work out the displacement? How far she found erbut? Yeah, I still remembering the solution at yesterday. We maybe we should we we should measure all the rectangle no, we should measure the square, the number of square and and in regular picture, I mean, Yeah so this fish is a nice big rectangle. We can do that. So that's 30 times ten. So 30 times six plus. This is a regular triangle. So that's half the base. So a half times two. Times 20. So we've got those two bits. We can get another bit of a rectangle here 20 times. H. And then it's just this bit we have to worry about. So what you were talking about we can use for this fish. Firstly, that is the square good. And secondly, one, two, three, four. Okay eight in regular picture, so I divided by two, four and 30 times six. There she. Yeah. So let's add up all our areas and we'll see how near their answer we get half times, two times, 20. So that's ten. So 20 times eight is 160. 30 times 60. Because 18 hundred. Yeah 30 times six. Five times six. What is that is five times six. Four and ten. That is just three Yeah if you take that bis 5:20 to five to 25. Yeah, five. So how about Oh, six maybe. So why is that six? From four to ten. And it's awkward. You can't see the graph. I should have reduced it for the lesson. And then we have our one, two, three, four, five, six, seven, nine, ten, eleven, twelve, 13, 14. So 14 times half squares, we have to work out what these squares. Stand for. So this square, what does that stand for? So it's 2.5. Times one. So 2.5, 2.5 times a half, times 14. 2.5 times point five times 14 is 17.5. So I get 17.5 for the this shape. 17.5. So 14 times a half, times 2.5, I worked at 2.5 times, one times a half for the part squares. So let me add up everything plus 30, plus 18 hundred, plus 160, plus ten. I get 22.17 km. Displacement area under the graph one squares 20m. 410 to 430m. Why are they getting something so different? I suppose they worked out this size square and then all the part squares. Let's see what meththey used. Number of squares is four, she 21 square is. One square is 10m, 2s times 5m. So. 42 squares, 12345, Oh Yeah. They've taken big squares and averaged it to about 42 squares. So one, one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, 13, 14, 15, miss, why not? We can calculate. Okay, let me coculate to it. 37 plus four equal to 41 and each one it's two times five. Ten. So each area is ten, yes, 410. So that is 10m displacement. So they said one, two, there are 41 squares. Yeah. So 41 times ten. That is area 410. Yeah, 410. Okay. Okay, a projectile is thrown at an angle to the ground at a certain time. The projectile has a horizontal velocity of 23m per second and a vertical velocity of -10.1m per second. What is the magnitude of the velocity of the projectile at this time? Of the projectile. So we're trying to find that resultant, aren't we? Speed at this time horizontal distance, so. Serum yes. 23 square plus 10.1 square. 25.1 25 maybe 25m Yeah 25.1m per second. Okay, the motion of a long jumper during a jumper similar to that of a projectile under gravity. The figure below shows the path of the athlete above the ground during a long jump from halfway through the jump at position a to position b. At which contact is made with the sand on the ground. The athlete is travelling horizontally at a. So you can see. Center of MaaS of the athlete. They hit the ground here. During this part of the jump, the center of MaaS of the athlete falls 1.2m. Calculate the time between positions a and b. Between point a and b. So we know that the distance. They fall. Or from that to from that point to that point. So they fall. So we know the acceleration. We know the acceleration is 9.81. Gravity so and the speed, so initial speed is zero, 1.2, equal to a half, 80 square a half times. So that is a root, root, 1.2 times two, divided by 9.81. So the time is not 0.5, not not not point 4649 0.49. That's. 1.2 times two equals divided by 9.81 equals square root. No point 49s, Yeah. The athlete is moving horizontally at a with a velocity of 8.5m per second. Assuming there is no air resistance, calculate the horizontal displacement of the center of MaaS from a to b. So essentially they're asking horizontally. So in horizontal, that is the constant speed, constant Yeah, constant speed. So no air resists in the horizontal displacement of the center of MaaS. So if the time is not point 49s, so 8.5 times, four point and 0.49. 4.2. So the athlete in the image above slides horizontally through the sand, a distance of 0.35m before stopping, calculate the time taken for the athlete to stop ing, assuming the horizontal component of the resistive forces from the sand is constant. So. They slide in the sand once they hit the sand. Time taken horizontally, no point. As blood to stop the horizontal component of the resistive sis constant distance not one, three, five before stopping. I don't know. So it's. It was that was initial velocity and v will be zero. So. P is equal to U plus at s is equal to U T plus a half 80 squared. So s is the distance is a half. V ma plus U over t, so U is 8.5, v is zero because they're stopping times t so point zero 82 of a second is the time it takes to come to a halt. Okay, Jackson, do you want to do more of these or do you want to do something a different topic tomorrow? Yeah. Well, I want to, I want to do the electricity again because I don't I'm not sure whether I can do very well in emf and potential difference. Okay, so tomorrow we look at emf and potential difference and we look at past paper questions. How they ask is in a paper. Okay, Jackson, I'll talk to you tomorrow. Bye bye.