13) A golfer hits a ball at 100 m/s with an angle of elevation of 30 degrees towards a green. The acceleration due to gravity is 9.8 m/s^2 [down]. There is no resistance. a) If the green is at the same height as the golfer then find how long the ball is in the air, and how far away the ball lands from where it was hit. Given: Gravity: 9.8m/s^2, Vi= 100 m/s What I've solved for: (x direction) di= 0m, df= 441.66 m, Vi= 86.6 m/s, Vf= 0m/s, a= 0 m/s^2, t=5.1s (y-dir) di= 0m, df=127.5m, Vi=50 m/s, Vf= 0 m/s, a= -9.8 m/s^2, t= 10.2s) I solved for the final x and y direction distances using trig laws. What I'm having trouble with: b) If the green is 10 m higher than the golfer then find out how long the ball is in the air, and how far away the ball lands. [You do not need to recalculate what you did in Part A] My Answer (which was wrong): Since initial and final velocity remain the same, the ball will be in the air for the same amount of time as Part A and should land only 10 m higher than that of Part A which is 137.5 m in the y-dir and still 441.66 m in the x-dir. What am I doing Wrong?