The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of 59 m/s2 from rest relative to the flight deck and launches the aircraft in a distance of 97 m measured along the angled takeoff ramp. If the carrier is moving at a steady 26 knots (1 knot = 1.852 km/h), determine the magnitude v of the actual velocity of the fighter when it is launched.

Explanation:

We shall find the final velocity of  aircraft with respect to aircraft carrier using the following relation.

v² = u² + 2as

v² = 0² + 2 x 59 x 97

v² = 11446

v = 107 m /s

velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h

= 48.152 x 1000 / (60 x 60) m/s

= 13.37 m /s

This velocity of aircraft carrier will be added to the velocity of aircraft .

So absolute velocity of aircraft = 107 m /s + 13.37 m/s

= 120.37 m/s

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

access to city

explanation: