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.
What equation can I use to get the maximum height?
You can find the Kinetic energy at point B (1/2 mv2^2). That will equal the change in PE to the top (mgh)
set them equal, notice the mass divide out.
h= 1/2 (v2)^2 /g
The 12.6 meters, and v has nothing to do with the problem, it is not needed.
So, am I supposed to figure out v2?
And what is g, in this case?
To determine the maximum height reached by the stone above point B, you can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2gh
Where:
vf = final velocity at the maximum height (which will be zero at the highest point)
vi = initial velocity at point B (v/2)
g = acceleration due to gravity (-9.8 m/s^2, assuming the stone is thrown on Earth)
h = maximum height above point B
So, you can rewrite the equation as:
0 = (v/2)^2 + 2(-9.8)h
Simplifying the equation:
0 = (v^2)/4 - 19.6h
Now, to solve for h, rearrange the equation to isolate h:
19.6h = v^2/4
h = (v^2/4) / 19.6
Therefore, the equation you can use to get the maximum height reached by the stone above point B is h = (v^2/4) / 19.6.