A projectile fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at an angle of elevation of 30 deg with the horizontal.

vertical problem
Hi = 150 m
Vi = 180 sin 30 = 90 m/s
v = Vi - g t = 90 - 9.81 t
v = 0 at top so t = 9.17 seconds to top
h = Hi + Vi t - (g/2) t^2

h at top = 150 + 90(9.17) - 4.9(9.17^2)
= 150 + 825 - 412 = 563 meters above ground at top after 9.17 s

now how long before ground?
falls from 563 meters high
v = - g T
0 = 563 - 4.9 T^2
T = 10.7 seconds falling
total time = t + t = 9.17 + 10.7 = 19.9 seconds
horizontal speed = 180 cos 30 = 156 m/s
so 156 m/s for 19.9 seconds ---> 3102 meters horizontal to target

KEEP UP

Given: 180m/s - initial velocity, 150m- height of cliff, 30° angle
Unknown: H- greatest elevation, R- horizontal distance until it strikes the ground
H=150m + (Vosin30°)² /2g
H=150 + (180m/s sin30°)² /2(9.81m/s²)
=562.844meters

R=(Vo² sin2(30°))/g
R=((180m/s)²sin(60°))/(9.71m/s²)
R=2860.267 meters