physics
posted by Antonio on .
A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 7.85 m higher than A, with speed v/2. Calculate (a) the speed v and (b) the maximum height reached by the stone above point B.

vf^2=Vi^2 +2gh
I assume you memorized that equation. Here we have.
(v/2)^2=v^22*g*7.85
solve for v.
Now, knowing the v, use the same equation as you started with (now Vf=0 at top, and vi=V/2, and you solve for h.