A plane travelling at a steady speed of 50ms-1 at an altitude of 300m releases a supply parcel when directly

above a point x on level ground. calculate
1) time of flight of the parcel
2)speed of impact of the parcel
3) distance of impact from point x

The velocity in the x direction will be 50 m/s
The time of flight can be found using the equation:
0 = 300 m - (1/2)(9.8 m/s²)t²
t = √{600/9.8} s
t = 7.82 s
The speed in the y direction is:
S = 9.8(7.82) = 76.68 m/s
The total speed is:
√{76.68² + 50²} = 91.54 m/s
The distance from point x is:
50(7.82) = 391 m

1)
from the equation h = 1/2gt^2 we get :
t = √2h/g = √600/9.8 = 7.825 secs
2)
Applying energy conservation principle :
1/2 m*Vo^2+m*g*h = 1/2m*V^2 we get :
Vo^2+2g*h = V^2
V = √50^2+600*9.8 = √8380 = 91.54 m/sec
with different approach :
Vy = √2*g*h = √19.6*300 = 76.68 m/sec
V = √Vo^2+Vy^2 = √2500+5880 = 91,54 m/sec

3)
d = Vo*t = 7.825*100/2 = 782.5/2 = 391.25 m