It will hit the ground when h(t)=0. Since the hammer is simply dropped its initial velocity, v0, is equal to zero and we are told that the initial height, h0, is 25m so our equation is:
h(t)=-4.9t^2+25, we want h(t)=0 so:
4.9t^2-25=0 add 25 to both sides
4.9t^2=25 divide both sides by 4.9
t^2=250/49, since t>0
t=+√(250/49) seconds
t≈2.3 seconds (to the nearest tenth of a second)
