PART A Let x(t) = 5 cos(t) ELLIPSE and y(t) = 3 sin(t) N At t = 6, -2 x 6 = -2 - y(6) speed(6) = PART B Now, if we let x(t) = 5 cos(2t) and y(t) = 3 sin(2t), we have the same path: ELLIPSE And at t = 3 x 3 = y(3) = Do you see how it's the same point as PART A? speed (3) Is speed in PART B slower or faster than the speed in PART A? Select an answer slower faster ost to forumAfter t seconds, a projectile hurled with initial velocity on and angle 0 from an initial height of ho will be at m(t) = ('00 cos(9))t feet and y(t) = ('00 sin(3))t 16f,L2 + ho feet. (This formula neglects air resistance.) initialspecd=v 9 3 initial angle (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 m(T) represent physically? O the time at which the projectile is at a maximum 0 the maximum height of the projectile O the total distance the projectile travels horizontally O the distance the projectile has traveled horizontally when it is at its maximum height (1 d (b) For m, 6 and ho in part (a), calculate _y and find T2 so that _y = 0 at t = T2. dm d2: What does m(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 O the distance the projectile has traveled horizontally when it is at its maximum height 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, 3:03) = t4 8223 + 16152 and y(t) = 7cos(t) for 0