Question

A balloon is ascending at the rate of 14 $m{s}^{-1}$ at a height of 98 m above the ground when the food packet is dropped from the balloon. After how much time and with what velocity does it reach the ground ? Take g = 9.8 $m{s}^{-2}$

Answer 1

Given,

Velocity of the balloon = initial velocity of the food packet $u=14\hat{j}m{s}^{-1}$

Gravity acceleration $a=-9.81\hat{j}m/s^2$

Distance from ground $h=-98\hat{j}m$

Final velocity $=v$ & time taken to reach the ground $=t$

Apply second kinematic equation

$v^2-u^2=2ah$

$v=\sqrt{2ah+u^2}$

$v=\sqrt{2(-9.81)(-98)+{14}^2}$

$v=-46\hat{j}m/s$

Apply first kinematic equation

$v=u+at$

$t=\frac{v-u}{a}=\frac{-46-14}{-9.81}=6.11\sec$

At velocity $46m/s$ and in time $6.11\sec$ it will reach to the ground.