Question

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?
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.
T=20s
what is the formula

Related Questions

A bomber flying horizontally at 250 m/s releases a bomb 1500m above the ground. At the point of impa... A bomber is flying horizontally over level terrain at a speed of 250 m/s relative to the ground and... A bomber in level flight, flying at 92m/s, releases a bomb at a height of 1950m. (a) How long is... A bomber, flying upwards at an angle 53 with verticql releases a bomb at 800m altitude. The bomb str...