Question

A balloon is ascending at the rate of $14m/s$ at a height of $98m$ above the ground a packet is dropped from the balloon. After how much time and with velocity does it reach the ground?

Answer 1

it will be the case of projectile thrown vertically up with a speed as velocity of balloon adds up with the packet , so in the case $u=14m/s,a=-g=-10m/s^2,s=-98m$,

We have to find time and velocity just before hitting the ground,

for time , we use 2nd eq of motion i.e, $s=ut+\frac{1}{2}a{t}^2$

$-98=14t-5t^2$

$\Rightarrow 5t^2-14t-98=0$

solving for the quadratic equation we get $t=4.8s$

And for the velocity , we use 1st equation of motion , $v=u+at$

$v=14-10\times 4.8=-34m/s$