A level fight bomber flying at 300 ft/s releases a bomb at an elevation of 6400 ft. Approximately how long before the bomb strikes the earth?

Answer 1

Only the vertical component counts for the time for descent.
Assuming no air resistance, we have from kinematics,
S=ut+(1/2)at^2
where
S=distance = 6400 ft
u=initial velocity (=0 ft/s vertical in this case)
a=acceleration due to gravity, g=32.2 ft/s²

Substitute values,
6400=(1/2)(32.2)t²

Solve for t.

Answer 2

T=20s

Answer 3

what is the formula