Question

A balloonist is ascending at a velocity of $12{\text{ms}}^{-1}$ and acceleration $2{\text{ms}}^{-2}$. A packet is dropped from it when it is at a height of $65\text{m}$ from the ground. Time taken by the packet to reach the ground( in seconds) is

Answer 1

Initial velocity of packet dropped from the balloon is same as velocity of the balloon.
So $u=12{\text{ms}}^{-1}$
Using $h=ut+\frac{1}{2}g{t}^2$ we get
$-65=12t-\frac{1}{2}\times 10\times t^2$
$5t^2-12t-65=0$
$(5t-13)(t-5)=0$
Solving t = 5 s or $-\frac{13}{5}s$
Time cannot be negative, thus t = 5 s.