Speed of stone thrown

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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?
 

Chi Meson

Science Advisor
Homework Helper
1,767
10
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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