Question

1) [25 pts] A golfer hits a ball at 210 ft/s and at an angle above horizontal of 12 degrees. The ball follows a projectile motion path with no aerodynamic drag. Two seconds after the ball has left the origin of the X-Y coordinate system determine: a) the ball's velocity vector in Cartesian coordinates INCLUDE units f b) the ball location coordinates in polar (r, e) coordinates 731 la) 205.4 c) the velocity component in the theta ( 2 ) direction d) the second time derivative of radial position coordinate (r) n r (c) Vo= 210 ft/s a = 32.23 €115² o 120 - 12

Answer 1

(a) Calculate x and y components of velocity of a ball initially as follows: U = u cos e = 210 x cos 12° = 205.41 fts U, = usin = 210 xsin 12° = 43.66 ft/s Calculate x and y components of velocity of a ball after two seconds as follows: V: = +at Along x direction acceleration is zero. v= 205.41+(0x2) = 205.41 ft/s Vy = U, +at =u, +(-8) = 43.66+(-32.2x2) =-20.74 ft/s Write the ball's velocity vector in Cartesian coordinates after two seconds as follows: v=vi+v, j = 205.411 +(-20.74)j = (205.4li – 20.74j) ft/s

(b) Calculate the position of the ball in x and y direction after two seconds as follows: =ut +-at- = (205.41x2) + (0x2) = 410.82 ft y=u,t +3a, 1 = (43.66x2) + (+32.242) = 22.92 ft coordinates after two seconds as Obtain the ball location coordinates in polar (r, e follows: r= Ver? + y2 = 410.822 +22.922 = 411.46 ft =tan( 22.92 tan 1 410.82 = 3.190 (r, 0) =(411.46 ft, 3.199)

(c) Calculate the velocity component of the ball after two seconds in the theta direction as follows: Vorro Apply derivative on both sides. (xy - řxy) (x2 + y ) = xj — *xy x=u+za1 =u_t+2(0xt") =ut i= =

y=u,t+, =u+(8) After two seconds, * = 205.41 ft/s y = 43.66-(32.2 x2) =-20.74 ft/s 0= (410.82*(-20.74)-205.41x 22.92) 411.462 = -0.078 va=411.46x-0.078 =(-32.09 ê) ft/s

(d) Calculate the second time derivative of radial position coordinate after two seconds as follows: p2 = x² + y2 Apply derivative on both sides. 2rr = 2x +2yj rr = x + yj Again, apply derivative on both sides. rr = xx + yj rï += (x8 + **) + ( yž + yy) xx+(8)?]+[vý +(sv)?]-(0) After 2 seconds, x =u, ru cos e dlo y =usin 0-gt

po2 = x² + y2 r = x + yj . xx+yj (410.82% 205.41)+(22.92x–20.74) 411.46 = 203.935 ft/s [410.82x0 +(205.41) 7+[(22.92x-32.2)+(-20.74)? 7-(203.935)? 411.46 =(0.719 ê,)ft/s?