An airplane with a speed of 74.5 m/s is climbing upward at an angle of 64.8 degrees with respect to the horizontal. When the plane's altitude is 687 m, the pilot releases the package. (a) Calculate the distance along the ground measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.
Vertical speed = 74.5 sin 64.8 = 67.40961541
Extra height reached = 4544.05625 / (2 x 9.8) = 231.8396046
Total height reached = 231.8396046 + 687 = 918.8396046 m
Time to reach max height = 74.5/9.8 = 7.602040816 s
Time to fall 919 m = square root (2 x 918.8396046 / 9.8) = 13.69373166 s
Total time in air = 21.29577248 seconds
Horizontal speed = 74.5 cos 64.8 = 31.72055722 m/s
Horizontal distance = 31.72055722 x 21.29577248 = 675.5137695 m
Vertical speeds at ground = 9.8 x 13.69373166 = 134.1985703 m/s
Angle to the ground velocity = tan^-1 (134.1985703 / 31.72055722) = 76.70106771 degrees
For part A my answer is 134.1985703 m/s
For part B my answer is 76.70106771 degrees
However, the answers are incorrect. What am I doing wrong? Can you please help. Thanks.