Kinematics with one variable

[~SPaRtaN~]

1. The problem statement, all variables and given/known data
A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 5.3 m higher than A, with speed v/2. Calculate the maximum height reached by the stone above point B.


2. Relevant equations
v(final) = v(initial) + at
x(final) = x(initial) + v(initial)*t + 1/2 at^2
x(final) - x(initial) = 1/2 [v(final) - v(initial)]*t
2a[x(final) - x(initial)] = v(final)^2 - v(initial)^2

3. The attempt at a solution
Vb^2 = Va^2 - 2 * g * Xab, where

Xab is given = 5.3 m
Va = V
Vb = V/2

(V^2) / 4 = V^2 - g * Xab

3/4 * V^2 = 2 * g * Xab

V^2 = 8/3 * 9.8 * 5.3 = 138.5 m^2/s^2
V = 11.77 m/s

Now use same equation between B-->C:

Vc^2 = Vb^2 - 2 * g * Xbc, where

Vc = 0
Vb = V / 2 = 11.77 / 2 = 5.88 m/s

solve for Xbc:

Xbc = Vb^2 / (2 * g)
Xbc = (5.88)^2 / (2 * 9.8)
Xbc = 1.76 m

So the height above point B is just

H = Xab + Xbc = 5.3 + 1.76 = 7.06 m



The above answer is not right.......help

[Kurdt]

3. The attempt at a solution
Vb^2 = Va^2 - 2 * g * Xab, where

Xab is given = 5.3 m
Va = V
Vb = V/2

(V^2) / 4 = V^2 - g * Xab

3/4 * V^2 = 2 * g * Xab

V^2 = 8/3 * 9.8 * 5.3 = 138.5 m^2/s^2
V = 11.77 m/s

Now use same equation between B-->C:

Vc^2 = Vb^2 - 2 * g * Xbc, where

Vc = 0
Vb = V / 2 = 11.77 / 2 = 5.88 m/s

solve for Xbc:

Xbc = Vb^2 / (2 * g)
Xbc = (5.88)^2 / (2 * 9.8)
Xbc = 1.76 m



I believe that should be the answer since the question asks for the max height beyond point B.

[~SPaRtaN~]

^ Thanks.........silly me, didn't paid attention on that part