Question

A balloon which is ascending at the rate of $12\text{m/s}$ is $80\text{metre}$ above the ground, when a stone is dropped. After what time the stone will reach the ground? [Take $g=9.8{\text{m/s}}^2$]

• A. $3\text{s}$
• B. $3.5\text{s}$
• C. $5.45\text{s}$
• D. $6\text{s}$

Answer 1

The correct option is C $5.45\text{s}$

Given, for the stone
inital velocity: $u_0=12\text{m/s}$,
initial height $h=80\text{m}=y_0$
Final height of stone H = 0 m =$y$
Now we need to find the time stone took to finally reach ground
From second equation of motion,
$y-y_0=u_{0}t-\frac{1}{2}g{t}^2$
$\Rightarrow 0-80\text{m}=\left(12\text{m/s}\right)t-\frac{1}{2}\left(9.8{\text{m/s}}^2\right){\text{t}}^2$
$\Rightarrow \left(4.9{\text{m/s}}^2\right){\text{t}}^2-\left(12\text{m/s}\right)t-80\text{m}=0$
$\Rightarrow t=5.45\text{s}$ or $t=-3.00\text{s}$
As time cannot be negative
$\therefore t=5.45\text{s}$