# Speed of stone thrown

#### Hyperreality

A stone is thrown vertically upward. On its way up it passes point A with speed v , and point B, 3.00 m higher than A, with speed 1/2 v . Calculate the speed v
Here is how I tackled this problem. Let v = final velocity, and u = initial velocity.

v^2 = u^2 - 2gd eq.1

v^2/4 = u^2 - 2g(d+3) eq.2

Rearrange eq. 2 to obtain
v^2 = 4u^2 - 8gd - 24

Solve u
4u^2 - 8gd - 24 = u^2 - 2gd

3u^2 = 6gd + 24

u^2 = 2gd + 8

v^2 = 2gd + 8 - 2gd
= 8
v = 2 [squ] 2 ms^-1

But the book says the answer is 8.85 ms^-1. So what am I doing wrong?

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#### Chi Meson

Homework Helper
There' only one equation.

When the stone passes point A, that's position "zero" with initial velocity "v". At point B, the displacement is now 3 m and final velocity is "v/2." You chose the right equation. Along the way to the answer you shoul get "(3v^2)/4 = 6g "

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