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(BcAp21) projectile fired horizontally from qun that 51.0 above flat ground: The muzle vcbaty projectile remain the air? 3.23 You ara correct: Previous Tries Your r...

Question

(BcAp21) projectile fired horizontally from qun that 51.0 above flat ground: The muzle vcbaty projectile remain the air? 3.23 You ara correct: Previous Tries Your recelpt no 153-9462 what horizontal distance from the firing point does strike the ground? 03*102 You correct: Previous Tres Your receipt no 153-0346 Whal #peed strikes the ground?nenuIncorrect: Tcles 615 Previous Trics

(BcAp21) projectile fired horizontally from qun that 51.0 above flat ground: The muzle vcbaty projectile remain the air? 3.23 You ara correct: Previous Tries Your recelpt no 153-9462 what horizontal distance from the firing point does strike the ground? 03*102 You correct: Previous Tres Your receipt no 153-0346 Whal #peed strikes the ground? nenu Incorrect: Tcles 615 Previous Trics



Answers

A projectile is fired at a height of 1.5 $\mathrm{m}$ above the ground with an initial velocity of 100 $\mathrm{m} / \mathrm{sec}$ and at an angle of $30^{\circ}$ above the horizontal. Use this information to answer the following questions:
Determine the range of the projectile.

So the question states that a projectile is shot at some angle theta at a speed V and travels 23 meters. And we're trying to find how far this project I will go. If we doubled the speed it was launched it but kept the angle the same. So the first thing we need to do is separate are, um, velocity vector into components. So we know this is going to be equal to be co sign data, and this will be equal to be signed data. And we get this because scientist Aita is opposite over hype on news. So opposite is, um, our visa by or this vertical velocity vector and are high pot news is V so that we could just solve for be supply. Same thing goes for Vico Sant data in height. Find it. Um, so now that we know this, we know that the range of the project I was going to be equal to the frosty times, the time evil, the delta tax and the velocity in the horizontal direction is going to be equal to the velocity Times co sign data and then this will be multiplied by the time which will give us our tell tax. So the main thing we need to do here is so for T in terms of Fada and V And to do this we can use our kinetic equations which state that the change in displacement in this case in the Y direction is equal to the initial velocity times a time plus for 1/2 times the acceleration time to the time squared. And so the we're told in the problem that the projectile starts on the ground and it ends on the ground. So its displacement is zero. We know its initial vertical velocity is the sign data. I don't know what tea is, and we can just leave acceleration as a We know it's going to be negative 9.8. But honestly, that is a and then t squared. And so from this we confected at a T. So get he signed Data plus one have a t all this most played by T and it's equal to zero. And so when we saw for tea will find that he is either equal to zero or a T is equal to negative two. The sign data over a So we can ignore the route that says that T is equal to zero because this doesn't make any sense in our problem. So we can use this route and plug it back into our equation up here. And when we do this, we find that the range is equal to, uh minus two times the velocity squared times sign. They, uh, Times co sign data all over the acceleration. So if we tailor this to our problem, let's say, uh, for the, um, the situation where the projectile only goes 23 meters, we'll call this Delta X one. And we'll also called this velocity V one. Um, and we can say this is also the one. It's still just make. It would be to be one anyways, So if we say Delta X one is going to be equal Teoh minus two times the velocity of the one square times signed data Times co sign data all over the acceleration. And if we look at the second situation, we can say that the, um, the displacement. So we'll call it Delta X two, And I'm actually gonna change this to, instead of to be one. I'll just make this be to so Delta X to the range of the velocity. The range of the projectile here is equal to minus two times feet too. Squared times signed data co sign data because the angles the same in both situations divided by a And so we confined the ratio between these two. So I'm gonna take dealt, uh, x two and divided by Delta X one. And when we do this will end up with the velocity of the second projectile squared, divided by the velocity of the first projectile squared. And this gives us a relationship where we confined. Um, how far this project I was gonna dio, given that the velocity is two times the original projectiles, boss T. So this V two is actually the one times two. So when we plugged that in, we see that Delta X two over Delta X one is equal to four v one squared over the one squared in This cancels and we see that Delta X two is equal to four times Delta x one, which means that the range of the second projectile is four times the range of the first projectile, which means it's going to be four times 23 meters, which is equal to 90 to meters. And that's the answer

This question. It is about projectile motion, so given. Okay, um, the range is 23 m with some you're not, and some they do not. Okay in. Now the the launch velocity Istanbul. Okay, so what happens to our Okay, So to do this question Oh, yeah. You can look at, uh, and use this formula. The range formula r equals two, you know, square sign to data. Not there are by chain. Okay, so, uh, she is constant. They don't know it's the same. Um, and so our will be proportional to not square. Okay, So are to divide by our one is equal to, you know, to divide by, you know, one square and so, uh, to we just be, you know, to develop a new one square times are one. Now, the you know to is double one. Okay, so we have to square are one four times 23. You get 92. So the, uh, the range increases right? Four times. Mm. Right. So, I Yeah, it kind of makes sense because we can also think of it as our it goes to the x component of the velocity times time when you double the launch. Ah, speed, you double the time. At the same time, it also double the X component of the velocity. So there are increases by four times, and that's all for this question.

We have all the number 53 In which 30 projectile Which is a little of the projectile motion or kind of takes in two and 3 dimensions. And uh from the topic projectile Okay There is a total which is fired with an initial speed of 46.6. So little supposed to use the initial speed 4 to 6.6 m per second. And angle theta is 42 points 2° like this. It is being fired. This is theta and this is you okay? No in part we have to find maximum height. So we have formula for maximum right. If this projectile goes like this, this is the maximum height and at maximum height we will be having only horizontal component. So we have formula for maximum height Which is Article two. Uh You Esquire Science Quality to Buy two G. So maximum height let us say aah equal to be squired Science Squad 3- two G. Science squatty to buy. So let us plug in the formula us squarish. Right? Because you have assumed you to be the initial velocity. Okay. 46.6 old square Science Choir 42.2° By two in 2 9.8. So we have to use computer for this for to 6.6 old square in two. Uh, saying for the well maybe we have to change it. Any radiant mud. Sorry, degree mount. Okay. 46.6 Questions Square 42.2 1942.2° was hold square into 46.6 Old Square. Do you want by two in 2 9.8. That is 49.99. That is 49.99. Approximately Maximum height is 50 meters. Okay, part pain by being is the total time in the air. So total time of flight. We need to find out the total time of flight. It will be to use scientology. Total time of flight to you Scientific to buy G. So this is two in 2 46.6. Sign 42.2° By 9.8. Again we have to find, we have to use a calculator for this. The sign 42.2° into 46.6 into two, Divide by 9.8. Total time of flight A 6.3881 for approximately six 3/9 seconds. This is a total time of flight. I think park si I've seen the total horizontal distance. So we have formula for horizontal distance which is you esquire scientific, horizontal distance. That is this distance which is also known as the range are equal to are equal to us. Choir sine two theta by G. You esquire signed to to to buy G. Okay so the C U. S square 46.6 whole square sign to into 42 Point What is this angle 42.2° 42.2°.. Bye 9.8. Yeah. Again to use computer. Same 84.4° In 246.6 called square Divided by 9.8. That is 220.53. So this is the oriental range meters of course 220 25 3 m/day. Mhm We have to find the velocity of the project Tile 1.50 seconds after fighting. So the velocity of the projectile after 1.5 seconds. So you know that the horizontal component at this velocity will be you cause theater and vertical component will be you scientist to. So after 1.5 seconds After 1.5 seconds if along why access it will be we were equal to U. Y. initially bless G and T. So initially in this direction. Very G. And in a positive action that is in vertical downward direction. So you while that initial velocity you sign you scientist or it could be you scientist to using data minus G. And this is and time GT. So let's plug in the values 46.6. Sign 42.2°- G. 9.8 in to 1.5 seconds. Yes, well We have to use that greater for this 46.6 In to sign 42.2 minus 9.8 into 1.5. This is 16.60 meter per second. Okay? No but we know that there is no exaggeration or there is no external force in our general direction. So uh you X equal to re X equal to always, always you go straight to You is 46.6 Costs 42.2°.. So again we have to thank you. 46.6 in two Course 42.2°.. There it is 34.52 14 You know the magnitude of this velocity, we will be the excess choir plus we were square And so we access quiet. We have 34.5 to 14 whole square bless Plus 16.6 16.60 Hold Square on the road. Okay, wow. Less 16.6 whole spread. And the dude that is 38.305. The velocity after 1.5 seconds approximately 38.31 meter per second. Thank you.

This problem we have given in the absence of air distance it projected is launched entrance to the ground level. So here it has range 23 m of us. This is any particles. It is projected with a speed we at angle theater and then it returns to the ground like this. This range is given to us 23 m. We know that in projectile range formulas are as a cool do we squared, signed to teeter waging So initial, this is the condition. And finally, if the launches spirits deviled and projectile is fired at the same angle of the ground What is the new range? So the new range will be equal to read a square scientific hitter by Jean Hear the prime is equal to twice of V What three days saying in this case I finally we call to four We square scientific literacy so we can say our prime will be called four times so far. So the new range will be equal to four and 2 23. That means yeah, this is a 92 meters


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