Jeff is standing on a bridge 5M above the river, He picks up a small rock and tosses it into the air, the rock reaches a height of 12M in one second and hits the water in about 2.3 seconds Determine when the rock reaches a height equal to the bridge but on the way down.

Answer 1

hf=hi+vi*t-1/2 9.8 t^2
vf=vi-9.8t at the top, vf=0, t=1
vi=9.8m/s
then
hf=hi+vi*t-4.9t^2
but hf=hi, so
4.9t^2-vi*t=0
t(4.9t-vi)=0 or t=9.8/4.9=2seconds
which is curious, and you need to think about it. It took just as long to go up as down...