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Homework Statement
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.
Homework 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
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