I need help with part 2 please. I have part 1 worked out.

MATH 2215 - 024 Project #2 October 9, 2019 Due : October 23, 2019 Part I : You are to fire a cannon and are trying to hit a target 300 meters due east of your current position. Your cannon fires projectiles with a mass of 2 kg and with an initial speed of 60 m / s. Gravity is pulling the projectile downward at an acceleration of 9.8 m / s2.Also there is a wind blowing from the north pushing the projectile southward with a force of 3 N. You need to decide two things. What is the angle 0 up from the ground you must aim the cannon and what is the angle @ the direction you must turn the cannon to the north. In the pictures above, the z direction is up, x is east and y is to the north. The first picture shows the situation from the side and the second picture shows the situation from above. Also there is a large ridge between the cannon and the target, so your angle of inclination 0 must be greater than 45 Use the acceleration vector given by the wind and gravity, the initial velocity, and the initial position (the origin) , to find a position function for the projectile. This will depend on t, 0, and p. Once you have this function, solve for t, 0, and p, using Solve, think about what the position vector should equal to when the projectile lands.) Make sure to plot out your solution. When you plot out the trajectory you may use the surface z 10000 / ((x - 150) 2 + 100) as the ridge. Calculate your angles to one tenth of one degree (remember Mathematica will be using radians) and make sure your projectile lands within one meter of the target in each direction (along x and y. ) You only have one shot, so make it count.Calc Project #2.pdf * + X O file:///C:/Users/sovit/Desktop/Calc%20Project%20#2.pdf ... G Google 2 of 5 - + Fit to page [ Page view A) Read aloud Add notes north Top view 1 3N - force due to wind West t = 300 m East South Mass of projectile, m = 2 kg So, the acceleration due to wind is aw = 1.5m / s? [towards the south] North I u cose East Type here to search O w 1:33 PM 10/19/2019Calc Project #2.pdf * + X O file:///C:/Users/sovit/Desktop/Calc%20Project%20#2.pdf ... G Google 3 of 5 - + 2 Fit to page [ Page view A) Read aloud Add notes X U = initial velocity Component of initial velocit along East is Vx Ux = UCos Sin [ ] Components of initial velocity along north is Vy Uy = UCos Sin [ ] Component of initial velocit along upward direction is Vz U2 = USin Acceleration along upward direction az = -g [gravity acting downward] Acceleration along East, ax = 0 m/ s? Acceleration along North, ay = 1.5m / s? [wind drag] Now, if x, y, z are the displacements from the origin along the corresponding axis at time t, then, x (t) i = y (t) j + z (t) K = = (t) [position of the particle] Now, we can use the formula of kinametics, s = ut + = ft2 component wise Where, s = displacement at time t u = initial velocity = acceleration So, x (t) = Uxt + = axt2 x (t) = UCos Coso t .. . . . equation #1 then, y (t) = Uz t + = at2 y (t) = UCos Sing t - 1.5t2 x _ = UCos [e] Sing t - 0.75 t2 . .. . . . equation #2 and, z (t) = Uz t + = az t2 = Using t - 2 t2 .. . ... . .. . . equation #3 Type here to search O w 1:34 PM 10/19/2019Calc Project #2.pdf * + X O file:///C:/Users/sovit/Desktop/Calc%20Project%20#2.pdf ... G Google 4 of 5 + 2 Fit to page [ Page view A) Read aloud Add notes Equations #1, #2 and #3 together will give the portion of the projectile at time ' t' At time t = T, when x (T) = 300 m then, y (T) = 0m and, z (T) = 0m From equation #3, T= = USin T = 2 USin [ ] . . . . . equation #4 Now, from equation #2, usin [6] Cos ...... ....equation #5 3 so from equation #4 and #5, - usingCose = 2 USine - U SingCose = 2 USine 3 tano Sing = = 2 y . . . . . equation #6 From equation #1, 300 = U Cos e Coso T = U Cose Cost 2U - Sin e = CoseSine Coso = 300 * 8 300 * 9.8 2 * 60 * 60 g = 9.8m / 5^2 U = 60 m / s 9.8 24 = CoseSine Coso = 2.8 24 . . . . . . in equation #7 = CoseSine V1 - Sing 9.8 = Cosesine 1 - 3 sin's 9.8 2g cos26 24 = Sine 1 - sin2 6 - > sine = (24) = (1+ 3 sino- sine+ 2.8=0 Type here to search O w 1:34 PM 10/19/2019Calc Project #2.pdf * + X O file:///C:/Users/sovit/Desktop/Calc%20Project%20#2.pdf ... G Google 5 of 5 - + Fit to page [ Page view A) Read aloud Add notes Calc Project #2.nb | 5 so , Sin 0 = 0.226, 0.641 = Sino = 0.475, 0.8 = 0 = 28.36 , 53.13 . substituting in equation # 6, $ = 4.739 , 11.77569- Now , g should take only the solution where e > 45 . [as said in the question] then, 6 = 53.13 - and $ = 11.7757 Now, form equation #4, T - 2 Usin _ 2 x60 x Sin [53.13: ] 9.8 = 9.796 Sec Therefore, the required time = 9.796 Sec. Part II : Shortly after you destroy the target your enemy will launch a reconnaissance plane along the ridge to survey the damage. The plane is automated and flies along a set path flight given by the curve : 10 r (t) = [150, 20 t - 300, 100 + 3 t -] In this function t is time in seconds after the plane is launched. Your goal is to shoot down the plane I Repeat the steps from above except this time you can assume the cannon is being fired due east so (p = 0) Now you need to determine the angle of inclination 0 and how many seconds after the plane is launched you should fire the cannon. Choose the smallest angle feasible to minimize the effect of things we are not taking into account (like wind resistance. ) Don't forget the wind is still blowing and you will need to take that into account. Again, calculate your angle to within one tenth of one degree and time to one tenth ofa second Note : The t in the path of the plane is not the same as the t in the equation for the projectile ! Type here to search O W 1:34 PM 10/19/2019