Question

A Balloon is moving upwards with a speed of 20m/s. When it is at a height of 14m from ground in front of a plane mirror in situation as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of sources S placed symmetrically before plane mirror during free fall.

Chapter:- Ray optics and optical instruments.
Class:- 12

Answer 1

$\longrightarrow$$\large\bold{\underline{\underline{\sf{\red{Question :-}}}}}$

A Balloon is moving upwards with a speed of 20m/s. When it is at a height of 14m from ground in front of a plane mirror in situation as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of sources S placed symmetrically before plane mirror during free fall.

$\longrightarrow$$\large\bold{\underline{\underline{\sf{\green{Answer :-}}}}}$

$\large\underline{\underline{\bigstar{\textbf{\textsf{\: Diagram \: is \: in \: attachment\::-}}}}}$

$\frac{FQ}{BF} = \frac{IH}{BH} = \frac{HS}{BH}$

$\therefore$

$FQ = (BF)( \frac{HS}{BH} )$

$(5)( \frac{1.0}{0.5} )$

10 m

$\dashrightarrow$FC = 2 + 10 = 12 m

v = 20 m/s

❍ Distance (s) = -12

❍ Initial velocity (u) = 20

❍ Time (t) = ?

❒ Second Equation Of Motion Formula :

$\begin{gathered}\begin{gathered}\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}\\\end{gathered}\end{gathered}$

★ Put values in given equation -

We will get -

➻ t1 = 4.535 sec

v = √(20)² - 2 × 10 × 10

$\large\underline{\blue{\underline{\frak{\pmb{ 14.14 \: m/s }}}}}$

$\begin{gathered}\begin{gathered}\mapsto \sf\boxed{\bold{\gray{v² - u² = 2gh}}}\\\end{gathered}\end{gathered}$

➻ t2 = 2v/g

$\large\underline{\orange{\underline{\frak{\pmb{ 2.83\: s }}}}}$

$\therefore$Time required = t1 - t2

$\qquad\qquad{\underline{\pmb{\purple{\mathcal{1.7 s}}}}}$

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