An airplane with a speed of 36 m/s is climbing upward at an angle of 54° with respect to the horizontal?

sin 54 = 0.809
cos 54 = 0.588
motion equation :
-620 = (36*0.809)*t-4.903t^2
-620-29.125t+4.903 t^2
t = (29.125+√29.125^2+19.612*628)/9.803 = 14.60 sec
distance d = Vox*t = (36*0.588*14.60) = 309.05 m
final speed V = √Vo^2+2gh = √36^2+1240*9.806 = 116.0 m/sec
Θ angle = arccos(36*0.588/116) = 79.5 below X axis

(a)
Consider the vertical direction :
Take g = 9.81 m/s²
Take all downward quantities to be positive, and downward quantities to be negative.

Initial speed, u(y) = -36 sin54° m/s
Displacement, s(y) = 620 m
Acceleration, a(y) = 9.81 m/s²

vertical direction: s = ut + (1/2)at²
620 = (-36 sin54°)t + (1/2)(9.81)t²
4.905t² - (36sin54°)t - 620 = 0
t = 14.6 s or t = -8.6 s (rejected)

Distance along the ground = (36 cos54°) (14.6) = 309 m


(b)
Let θ be the angle of the velocity vector of the package.

tanθ = 620/309
Then, θ = tan⁻¹(620/309) = 63.5°