y naught amounts to zero. The speed of the arrowhead is 35 meters per 2nd and us don't understand what the angle of launch is, but that's component A, is to number that out. Now because we have this convenient fact that the launch elevation of the arrowhead is the exact same as the target, we are permitted to usage this variety formula. We have the right to solve it for theta and then the will offer us this edge of launch. Therefore the range which is 75 meters, equals the early velocity of the arrowhead which is 35 meters every second, time sine of 2 times the start angle, all separated by the acceleration as result of gravity. Us will solve this because that sine 2 theta first by multiply both sides by g over v naught squared and g end v nothing squared cancels on the right and also it shows up on the left multiplied by r. Climate we'll move the sides around so we have actually sine two theta equals rg end squared. Climate take the inverse sine that both sides and on the left side that pipeline us with two theta and on the appropriate side we have actually inverse sine of rg over v naught squared. Then division both political parties by two and we have that our launch angle is inverse sine that the range, times acceleration because of gravity divided by the initial velocity squared, anywhere two. Therefore it's inverse sine that 75 time 9.8 end 35 meter per 2nd squared. By the way, whenever you view g it's always positive 9.8. This g is no the acceleration per se, yet it's just the magnitude of the acceleration because of gravity. Then this offers us 18.4 degrees above the horizontal is the start angle. Component B says that over there is a tree v a branch the is situated halfway between the launch place of the arrow and also the target. So this halfway point x subscript t because that tree, is fifty percent of the range x. For this reason that's half of 75 i m sorry 37.5 meters. We're told the the branch is at a elevation three and a fifty percent meters above the launch elevation of the arrow. Will the arrow go over the branch or under the branch is the question. For this reason that's another way of questioning at x equals 37.5, will y, the vertical position of the arrow, be higher than or much less than 3.5? so we have actually to discover out what is the y place of the arrow at this x place 37.5. For this reason our strategy will certainly be to very first find the moment at i m sorry the arrowhead reaches this x position, 37.5 and also then understanding that time, we'll then plug right into this formula for the elevation at that time. Therefore horizontally, we have that the x position is -- the early stage x place which is just zero, plus the x ingredient of the velocity times time and also we recognize what this is, and also we understand what this is and also so we'll fix this for t. The x component of the velocity is the velocity times cosine the theta and we're making use of cosine since the x component is in the horizontal direction. The is the surrounding leg the this triangle offered the angle is in here. Okay. We'll solve this because that t by dividing both political parties by v naught times cos theta and also then move the political parties around. We have t is x over v nothing cos theta. That's 37.5 meter the horizontal position of the tree, split by 35.0 meters per second times cosine that the angle the we determined in component A, 18.4349 degrees. I might have uncovered some other means to carry out this concern without using the number that we calculated in component A however it would have actually been quite facility looking. For this reason we'll simply make things a many simpler and also use the number anyway. Usually we would want to stop using calculated numbers to prevent propagating part error that we may have made prior. Therefore this provides us a time that 1.1294 seconds, keeping lots of digits there in bespeak to avoid intermediate rounding error. Currently the y position then at that time is walking to it is in the early yposition which is zero, add to the early y ingredient of velocity time time, add to one fifty percent times the y component of the acceleration, time time squared. This is the acceleration the gravity. The y component of velocity will be the early velocity time sine the theta since it is the opposite foot of this triangle. So that is 35 times sine of 18.4349 degrees times the time we figured out before, 1.1294 seconds, add to one fifty percent times acceleration of an unfavorable 9.8 meter per 2nd squared, time 1.1294 secs squared. That gives 6.25 meters will certainly be the elevation of the arrowhead when that is at an x place of 37.5 meters. That is more than the height of the branch the 3.5 meters so ours conclusion is that the arrow will walk over the branch.">


You are watching: An archer shoots an arrow at a target that is a horizontal distance

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This is university Physics Answers with Shaun Dychko. This archer fires an arrow at a target which is 75 meter away and the height of the target is the exact same as the launch height of the arrow, and we"ll say the is y naught equals zero. The speed of the arrow is 35 meter per 2nd and us don"t recognize what the angle of beginning is, however that"s part A, is to figure that out. Now since we have actually this convenient truth that the launch height of the arrowhead is the exact same as the target, us are permitted to usage this selection formula. We have the right to solve it because that theta and also then that will offer us this angle of launch. For this reason the variety which is 75 meters, equals the initial velocity that the arrow which is 35 meters every second, time sine of two times the beginning angle, all divided by the acceleration as result of gravity. Us will fix this because that sine two theta very first by multiply both sides by g end v naught squared and also g end v nothing squared cancels top top the right and also it shows up on the left multiplied by r. Then we"ll move the sides roughly so we have actually sine two theta equates to rg end squared. Then take the station sine the both sides and also on the left next that pipeline us v two theta and on the appropriate side we have actually inverse sine the rg end v nothing squared. Then division both political parties by two and we have actually that our launch angle is station sine the the range, times acceleration due to gravity divided by the early velocity squared, everywhere two. For this reason it"s train station sine the 75 times 9.8 over 35 meters per 2nd squared. By the way, anytime you view g it"s constantly positive 9.8. This g is not the acceleration every se, but it"s just the magnitude of the acceleration due to gravity. Climate this gives us 18.4 degrees over the horizontal is the start angle. Part B states that there is a tree with a branch that is situated halfway between the launch position of the arrow and also the target. So this halfway suggest x subscript t because that tree, is fifty percent of the range x. For this reason that"s half of 75 i m sorry 37.5 meters. We"re told the the branch is at a elevation three and also a half meters above the launch elevation of the arrow. Will certainly the arrow go end the branch or under the branch is the question. Therefore that"s another method of questioning at x equals 37.5, will y, the vertical position of the arrow, be higher than or less than 3.5? therefore we have actually to uncover out what is the y position of the arrowhead at this x position 37.5. So our strategy will be to an initial find the time at which the arrowhead reaches this x position, 37.5 and then discovering that time, we"ll climate plug into this formula because that the height at that time. For this reason horizontally, we have that the x position is -- the initial x place which is simply zero, add to the x component of the velocity time time and we recognize what this is, and we recognize what this is and so we"ll settle this for t. The x component of the velocity is the velocity times cosine of theta and also we"re utilizing cosine because the x ingredient is in the horizontal direction. It is the surrounding leg that this triangle given the edge is in here. Okay. We"ll fix this because that t by dividing both political parties by v naught times cos theta and also then switch the political parties around. We have t is x over v nothing cos theta. That"s 37.5 meters the horizontal position of the tree, split by 35.0 meter per 2nd times cosine of the angle the we established in part A, 18.4349 degrees. I can have found some other method to execute this inquiry without making use of the number that we calculated in component A but it would have been quite complicated looking. For this reason we"ll simply make things a lot simpler and also use the number anyway. Normally we would want to avoid using calculated numbers to avoid propagating some error that we may have made prior. Therefore this gives us a time that 1.1294 seconds, maintaining lots of digits there in bespeak to protect against intermediate round off error. Currently the y place then at the time is walk to be the initial yposition i m sorry is zero, add to the early stage y component of velocity time time, add to one fifty percent times the y component of the acceleration, times time squared. This is the acceleration that gravity. The y ingredient of velocity will certainly be the early stage velocity times sine that theta because it is the opposite foot of this triangle. So the is 35 time sine that 18.4349 degrees times the time we identified before, 1.1294 seconds, to add one fifty percent times acceleration of an adverse 9.8 meters per 2nd squared, times 1.1294 seconds squared. That provides 6.25 meters will certainly be the height of the arrow when the is at an x place of 37.5 meters. That is much more than the height of the branch the 3.5 meters so ours conclusion is that the arrowhead will walk over the branch.
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Is the not feasible to use Tan legislation in this equation offered that you have actually the angle opposite to the side(Y) and the nearby side (37.5m) ?
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For part B, rather of doing all those prolonged equations and also using the teta which us calculated i.e could be wrong, we can directly find the maximum height by using this formula: h= V nothing y Square/2g i.e the formula acquired from v_y^2=v_oy^2 + 2 g (y-y_o)
Hello SayedAbdul, you"re fairly right. Give thanks to you for pointing the out. This strategy is correct as result of the fortunate coincidence that the branch is halfway between the archer and the target, at which point the vertical component the the arrow"s velocity is zero. Had the branch been in a various horizontal position, climate the strategy shown would have to be used, or your strategy would need to be modified by detect the vertical component of velocity at the branch"s horizontal position. I"ll make a note around this in the final answer.All the best,Shaun
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