Asked by Srecko
An airplane traveling horizontally at 367 km/hr at a height of 1 km releases a bomb. How far ahead (to the nearest meter) from the release point does the bomb strike the ground?
Answers
Answered by
bobpursley
time to fall 1000 m: d=1/2 g t^2
t=2d/g=2000/9.8 seconds
367km/hr=367/3.6 m/s
d=v*t=367/3.6 * 2000/9.8 meters
t=2d/g=2000/9.8 seconds
367km/hr=367/3.6 m/s
d=v*t=367/3.6 * 2000/9.8 meters
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