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## Homework Statement

A man let a stone from height 49 metres. After 1 second he threw a second stone. Both stones fell on same time. What is the velocity of second stone?

## Homework Equations

h=1/2gt^2

h=1/2a(t-1)^2

## The Attempt at a Solution

So I assume that if he let the first stone then the acceleration is g

The second stone is thrown which means that his acceleration isn't g but a

I solved from equation h=1/2gt^2 the time and placed it into the next equation h=1/2a.(t-1)^2

I solved acceleration and put it into this equation v^2 = 2.a.s

I think the answer should be 45m/s but the answer in the book is 12m/s. How is this possible? Even the first stone won't fall with velocity 12m/s or am I wrong ?

I appreciate your help. Thank you