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# AB-Question 7

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### AB-Question 7 - Page Text Content

FC: BY ADAM BIRMINGHAM | QUESTION 7: Artillery on the Ridge

1: Question: 7 | Artillerymen fortified on a hillside are trying to hit an enemy on the other side of the river. Their cannon is at (x,y)=(3,250), where x is in kilometers and y is in meters. The target is at (x,y)=(-2,50). In order to avoid hitting the mountain on the other side of the river, the projectile fired from the cannon must go through (x,y)=(-1,410) | 1) Write the quadratic equation of the parabolic path of the Projectile 2) How high above the river will the projectile be where it crosses the right river bank at x=2? 3) How high above the river will the projectile be where it crosses the left river bank at x=0? 4) Approximately where will the projectile be at y=130? 5) A reconnaissance plane flying at 660 meters above the river. Is it in danger of being hit?

2: Quadratic Equation of the Parabolic Path | -80x^2+120x+610=0 | In order to find the quadratic equation for the parabolic path of the projectile, the first step is to place the values into their respective L1 (x) and L2 (y) values in order on your calculator. This is done (on the TI-83 plus, at least) by pressing the stat button followed by edit. Enter the After you finish entering the values the L1 column should read in this order: 3, -2, -1. L2 values should read as follows: 250, 50, 410. Next you click Stat again, go to the CALC menu and select QuadReg. This will bring you to the equation screen. Enter L1 and L2 (Separated by commas) following QuadReg and hit enter, if done correctly, the equation below should be derivable.

3: How high above the river will the projectile be where it crosses the right river bank at x=2? | The quickest and easiest way to find out the answer to this question is to access the table on the calculator.This is done simply by hitting 2nd then graph. Once there, scroll down the x column until you reach x=2, the height of the projectile at this point is equal to the corresponding Y value directly to its right. Doing such should provide you with the coordinates given below: | (2,530) | If this is in fact the case, when x=2, the projectile is 530 meters above the river.

4: How high above the river will the projectile be where it crosses the left river bank at x=0? | This can be answered by using the exact same process utilized on the previous page. The only significant difference is that height of the projectile is determined by the y value corresponding with x=0 as opposed to x=2. After making these slight adjustments you should come up with the following coordinates: | (0,610) | According to the table at this point the projectile is 610 meters above the river.

5: Approximately where will the projectile be at y=130? | Unlike the previous two questions, this cannot be answered by referring to the table as 130 does not appear in the Y column, meaning that it does correspond with an x value that is an integer. Instead make Y2=130 and use the CALC intersect function to find the two places where it intersects the trajectory. | (-1.81,130) and (3.12, 130) | Since 3.12 is behind the location of the artillery piece and 130 is 120 meters below it, it can safely be assumed the desired coordinates (-1.81,130)

6: A reconnaissance plane flying at 660 meters above the river. Is it in danger of being hit? | The plane is not in danger of being hit, this can be confirmed by repeating the process used to solve the previous problem. Place 660 in the Y2 value and using the CALC intersect function. This will cause an error to occur as the to trajectories never intersect, with the maximum height reached by the projectile is 650 meters.

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