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OpenStudy (anonymous):

A jet with mass m = 8 × 104 kg jet accelerates down the runway for takeoff at 1.5 m/s2. I have Found out so far that... 1) net horizontal force(takeoff) = 120000 2)Net vertical force(takeoff) = 0 Now... Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 20 m/s, while the horizontal speed increases from 80 m/s to 96 m/s. 3)What is the net horizontal force on the airplane as it climbs upward = 64000 I need help here... 4) What is the net vertical force on the airplane as it climbs upward?

OpenStudy (plainntall):

The net vertical force for a 8x10^4kg jet that accelerates from 0 m/s to 20 m/s over 20 seconds is F=ma=m(V2-V1)/(t2-t1) there V2 is the final vertical speed and V1 is the initial vertical speed and t2 is the final time and t1 is the initial time. =(8x10^4kg)(20m/s - 0 m/s)/20s =(8x10^4kg)(1m/s^2) = 8x10^4 Newtons

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