Given:
u = 273 m/s, horizontal launch speed
h = 3.27 km = 3270 m, altitude.
v = 0, vertical launch speed
g = 9.8 m/s².
Wind resistance is ignored.
The time, t, for the bomb to reach the ground is given by
0t + (1/2)gt² = h
4.9t² = 3270
t² = 667.3469
t = 25.833 s
The horizontal distance traveled is
273*25.833 = 7052.4 m = 7.0524 km
Answer: 7.05 km
