A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 12.6 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B.

Question

I was told the equation to this was "h=1/2(v2)^2/g."
However, I’m still confused…do I have to solve for v2? And what would g be, in this case?

Answer 1

OHHHH. I thought you typed v2, now I see it at v/2. Big difference.

vfinal=vinitial -g t
v/2 = v-gt
t=v/2g

Now, use that time in the height equation.

12=v*t - 1/2 g t^2
12=v(v/2g) - 1/2g (v/2g)^2
now solve for v. Thence, you know v/2.

vf^2=vi^2 +2gh
g= -9.8 vf=0 vi= v/2 from above
h is the height it goes above B

Answer 2

Now, when I'm solving for time, do I not yet know what v/2 is? Like, it would just be "t = v/2 x -9.8?"

I'm so sorry...this is just not making any sense to me yet.