An airplane with a speed of 96.8 m/s is climbing upward at an angle of 74.0 ° with respect to the horizontal. When the plane's altitude is 906 m, the pilot releases a 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.
The Attempt at a Solution
I solved for part a. The distance x is 695.78m
I now need to determine the angle of the velocity vector of the package just before impact. I'm having trouble with this part, here is what I have so far:
v2y = voy + 2(ay)y
v= square root of vx2 + vy2 vox=vx vx=26.68
theta = 53.43 this answer is wrong I am not sure where I went wrong