An airplane with a speed of 70.6 m/s is climbing upward at an angle of 40.0 ° with respect to the horizontal. When the plane's altitude is 814 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.
70.6 * cos(40) = 54.1 velocity initial in the x direction = velocity final
70.6 * sin(40) = 45.4 velocity initial in the y direction
Displacement(Y) = -814 meters (initial to final position)
The Attempt at a Solution
I already obtained the answer for part A. What I did was solve for t using
-814 = 45.4*t - 4.9t^2
t = 18.3 = time of flight
so for part a) the horizontal distance = 18.3 * 54.1 = 990 meters
now for part b) I have the final velocity in the x (54.1), I need the final velocity in the y, and I can use inverse tangent to find the angle once v_final(y direction) is found.
To find this value I used v_y = v_initial(y) - g*t using 45.4 for v_initial(y) and 18.3 for time
which gives v_y = -133.94 and so theta = inverse tangent ( 133.94/54.1) = 68 degrees from the ground, however this is wrong .
I'm assuming my value for the final velocity in the y direction is wrong...