\fPART A Let 93(15) = 5 0080;) ELLIPSE and y(t) = 3 sin(t) 0\" y(20) = 3sin(20) speed(20) = C] PART B Now, if we let 23(t) = 5 cos(5t) and y(t) = 3 sin(5t), we have the same path: ELLIPSE 31(4): 5cos(20) s/ 0\" y(4)= 3sin(20) d\" " Do you see how it's the same point as PART A? speed(4) = C] Is speed in PART B slower or faster than the speed in PART A? ~/o' After t seconds, a projectile hurled with initial velocity 110 and angle 9 from an initial height of he will be at 93(t) = (170 cos(9))t feet and y(t) = ('00 sin(6))t 161%2 + ho feet. (This formula neglects air resistance.) initialspood=v 9 as initial male (a) For an initial velocity of 104 feet per second at an angle of E and an initial height of 1 feet, find T > 0 so that y(T) = 0. What does 93(T) represent physically? O the time at which the projectile is at a maximum 0 the maximum height of the projectile C) the total distance the projectile travels horizontally O the distance the projectile has traveled horizontally when it is at its maximum height 04 d d (b) For '00, I9 and ho in part (a), calculate % and find T2 so that (1: = 0 at t = T2. What does a: T2 represent physically? O the time at which the projectile is at a maximum 0 the total distance the projectile travels horizontally O the maximum height of the projectile Q) the distance the projectile has traveled horizontally when it is at its maximum height 04\\/ c) What initial velocity is needed so a projectile hurled at an angle of g from an inital height of 1 feet will go over a 43-foot high fence that is 480 feet away? You need an initial velocity of at least :] feet per second. For the parametric function, m(t) = t4 8153 + 16t2 and y(t) = 7cos(t) for 0