Kinematics problem – Bullet in block of wood

1. The problem statement, all variables and given/known data
A 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. What was the speed of the bullet immediately before the collision?

The attempt at a solution
d=1/2at^2
t=*sqrt(2d-vertical-)/a
t=*sqrt[(2)(1.75m)/(9.81m/s^2)
t= 0.597309633s

d= vt
vf = d(horizontal)/t
vf= 1.25m/0.597309633m/s^2
vf= 2.09216961m/s

Conservation of momentum (P1+P2)before = (Psystem) after
(m of bullet * vi of bullet) + (m of block * vi of block) = (m of system * vf of system)
(0.012kg*vi) + (0) = (3.512kg*2.09216961m/s)
(0.012kg)vi = 7.349621967kg*m/s
vi = 7.349621967kg/.012kg
vi = 612.4684m/s

The initial velocity seems a bit high and I’m not sure how else to approach the question. Help please!! 🙂

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