Question

A jet fighter lands on an aircraft carrier. m = 35,000 kg, the force of deceleration is 43,000 N, and the initial velocity is 27 m/s. What distance does it travel during the first 7.5 seconds of deceleration?
F=ma,
a=F/m,
s=v^2/2•a=m•v^2/2•F=
=35000•27^2/2•43000=296.7 m

Related Questions

A 16,000 kg F-18 (jet fighter) lands at 52 m/s on an aircraft carrier, its tail hook snags the cable... 2) An aircraft lands on a runway at a speed of 40 meters per second (ms^-1) and brakes to a halt in... A fighter jet lands on the flight deck of an aircraft carrier that has a length of 300.0 m. The jet... The aircraft carrier was first invented in 1914 during World War I, but the modern look of a carrier...