An object while in flight has an acceleration of a(t)= (0,0,-32). The initial velocity of the object is v(0)=(8,4,16) and initial position r(0)=(1,2,0). At what time t does not equal 0 and what position does the object cross the xy plane (this would represent where the object would hit the ground). How far did the object travel during the flight? Solution 1.x=8t+1 y=4t+2 z=16-16*t^2 crossing xy plane==>z=0 so t=1 2.at t=1 x=9 y=6 z=0 distance=sqrt(64+16)=8.944.