Question

A balloon is at a height of 81m and is ascending upwards with a velocity of 12 m/s. A body of 2kg weight is dropped from it. If $g=10m{s}^{-2}$ the body will reach the surface of the earth in how much time?

Answer 1

Given Data:

For balloon h(s)=81 m

So, body is also dropped from same hight of $81m$
Initial velocity, $\text{u=0 m/s}$

Final velocity, $\text{v= 12 m/s}$

Since body is dropped from ballon so it will also same velocity as ballon of, $u=-12m{s}^{-1}$
for a body acceleration due to gravity, g=10 m/s²
Step 1: Time taken by body to reach the surface:

$s=ut+\frac{1}{2}g{t}^2$

$81=(-12)t+\frac{1}{2}\left(10\right)t^2$

$81=(-12)t+5t^2$

$5t^2-12t-81=0$

$\text{5t²+15t-27t-81=}0$

$\text{5t(t+3)-27(t+3)=0}$

$\text{(5t-27)(t+3)=0}$

Time is positive so, $t=\frac{27}{5}=5.4sec$