Question

A balloon is moving upwards with a speed of $20\text{m/s}$. When it is at a height of $14\text{m}$ from the ground in front of a plane mirror as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of source $S$ placed symmetrically before the plane mirror during free fall. Take $g=10{\text{m/s}}^2$.

• A. $4.5\text{s}$
• B. $3.7\text{s}$
• C. $1.7\text{s}$
• D. $2.7\text{s}$

Answer 1

The correct option is C $1.7\text{s}$

Suppose $C$ is the highest point at which the image of source $S$ is visible and $G$ is the lowest point for which it is visible. Drawing the ray diagram for these two points,


From geometry, in $\Delta AHS$ and $\Delta ACD$

$\tan \theta =\frac{1}{0.5}=\frac{x}{5}$
$\Rightarrow x=10\text{m}$

$\Rightarrow CD=FG=10\text{m}$ [by symmetry]

Boy will see the image of source when it is in field of view from $C$ to $G$.
At point $A$, when he leaves the balloon,
initial speed, $u=20\text{m/s}$ [same as balloon]

Time taken to go from $D$ to $C$ is given by,
$s_C-s_D=10=20t_1-\frac{1}{2}g{t}_1^2$
$\Rightarrow 10=20t_1-5t_1^2$
On solving it, we get, $t_1=2-\sqrt{2}=0.59\text{s}$ ..... (1)

Time taken to fall from $D$ to $G$ is given by
$s_G-s_D=12=20t_2+\frac{1}{2}g{t}_2^2$
Solving the equation,
$\Rightarrow t_2=0.53\text{s}$

$\therefore$ The total time duration for which the image is visible will be
$t=2t_1+t_2=1.7\text{s}$