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2.Motion in Straight Line
hard
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
A
$\sqrt {\frac{h}{{8g}}} $
B
$\sqrt {8g\,h} $
C
$\sqrt {2g\,h} $
D
$\sqrt {\frac{h}{{2g}}} $
Solution
(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}}} $
Standard 11
Physics
