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?