The velocity of rock as a function of t {\displaystyle t} is v ( t ) = h ′ ( t ) = 10 − 4 t {\displaystyle v(t)=h'(t)=10-4t} .
The time that rock hits the ground: let h ( t ) = 0 {\displaystyle h(t)=0} , then 10 t − 2 t 2 = t ( 10 − 2 t ) = 0 → t = 0 , 5 {\displaystyle 10t-2t^{2}=t(10-2t)=0\rightarrow t=0,5}
t = 0 {\displaystyle t=0} is the moment that rock starts, t = 5 {\displaystyle t=5} is when rock goes back to surface.
v ( 5 ) = 10 − 4 ∗ 5 = − 10 {\displaystyle v(5)=10-4*5=-10}
answer: − 10 m / s , the negative sign gives the direction as downward {\displaystyle \color {blue}-10m/s,{\text{the negative sign gives the direction as downward}}}
