A pilot of an aeroplane flying at 300km\hr wishes to release a package of mail at the right position so that it hits the spot. What angle (theta) should his line of sight to the target make the instant of release at 150 metre.

Answer 1

time to fall 150m
h=1/2 g t^2
t=sqrt(2h/g)=you do it

distance from target to drop:
300,000m/3600sec=velocity figure that v.
then distance to drop=v*time above

tanTheta=distance/150

check my thinking