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**Motion Equation with constant speed (Time sensitive!!)**

## Homework Statement

A stone thrown horizontally from a height of 6.02m hits the ground at a distance of 12.90m. The initial speed is 11.64 m/s. Calculate the speed of the stone as it hits the ground. Neglect air resistance.

## Homework Equations

v^2-u^2=2as

v^2=2as+u^2

## The Attempt at a Solution

u (initial velocity): 11.64 m/s

a (acceleration): 9.81 m/s^2

s (displacement): 12.9-6.02=6.58 m

v^2=2*9.81*6.58+11.64^2

v^2=264.59

v=16.27 m/s

And that is not right? Any hints or helping suggestions?

Thanks!