Asked by Angel
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?
Answers
Answered by
MathMate
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.
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.
Answered by
Anonymous
T=20s
Answered by
Drammy
what is the formula
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