A rock is thrown upward with a velocity of 23 meters per second from the top of a 25 meter high cliff, and it misses the cliff on the way back down. when will the rock be 11 meters from the water, below
Vf=vi+at -23m/s=23m/s-9.8m/s^2(t) t1 from throwing up back to starting position t1=4.693s Delta(x)=vi(t)+.5a(t^2) 25-11=14 -14m=-23m/s(t)-4.8m/s^s(t^2) t2 is from 25m to 11 m t2=.546s Total time =5.239s
