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STD 11 - 2. motion in straight line — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 2. motion in straight line4 Marks
MCQ
A stone is dropped from a height $h$. Simultaneously, another stone is thrown up from the ground which reaches a height $4 \,h$. The two stones cross each other after time
  • ✓
    $\sqrt {\frac{h}{{8g}}} $
  • B
    $\sqrt {8g\,h} $
  • C
    $\sqrt {2g\,h} $
  • D
    $\sqrt {\frac{h}{{2g}}} $
✓

Answer

Correct option: A.
$\sqrt {\frac{h}{{8g}}} $
a
(a) For first stone $u = 0$ and

For second stone $\frac{{{u^2}}}{{2g}} = 4h \Rightarrow {u^2} = 8gh$

$\therefore u = \sqrt {8gh} $

Now, ${h_1} = \frac{1}{2}g{t^2}$

${h_2} = \sqrt {8gh} t - \frac{1}{2}g{t^2}$

where, $t =$ time to cross each other.

$\because {h_1} + {h_2} = h$

==> $\frac{1}{2}g{t^2} + \sqrt {8gh} t - \frac{1}{2}g{t^2} = h$ ==> $t = \frac{h}{{\sqrt {8gh} }} = \sqrt {\frac{h}{{8g}}} $

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