# What is the final velocity of a stone dropped out of a rising balloon? [closed]

A balloon is starting from the ground has been ascending vertically at uniform velocity for 4 sec and a stone let to fall from it reaches the ground in 6 sec. Find the velocity with which the stone hits the ground. ($g= 10 \frac{m}{s^2}$)

The answer given in my book is $42 \frac{m}{s}$ but I got $-60 \frac{m}{s}$ using the equation $v = u + at$. If my answer is wrong please tell why this equation didn't work.

Working :-
$t = 6 s$
$u = 0 \frac{m}{s}$ (left fall means stone is released from rest without any push)
It's under free fall $a = -10\frac{m}{s^2}$
$v = u + at$
$v = 0 + (-10)6$
$v = -60 \frac{m}{s}$

## closed as off-topic by Danu, Kyle Kanos, ACuriousMind♦, Prahar, John RennieOct 4 '14 at 14:42

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• could you please show your complete working? – user60063 Oct 4 '14 at 12:23
• Working added in question. – pcforgeek Oct 4 '14 at 12:46
• What is the 'uniform velocity' with which the balloon is moving? – Gaurav Oct 4 '14 at 13:19
• We have to assume it as u and we get initial velocity of of stone w.r.t ground which is equal to u. Then h = 4u. Substituting these value in 2nd equation of motion with constant acceleration we get u = 18 m/s. We get v = -42 m/s from v = u + at. – pcforgeek Oct 4 '14 at 13:55