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Standard XII

Physics

Motion Under Gravity

Two particles...

Question

Two particles of masses $m_{1} a n d m_{2}$ are dropped from height $h_{1} a n d h_{2}$ . They reach the earth after times $t_{1} a n d t_{2}$ respectively, Then:

\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}

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\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{2}}{h_{1}}}

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\frac{t_{2}}{t_{1}} = \frac{h_{2}}{h_{1}}

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none of these

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Solution

The correct option is A $\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$
$h_{1} = \frac{1}{2} g t_{1}^{2} a n d h_{2} = \frac{1}{2} g t_{2}^{2}$
$\Rightarrow t_{1} = \sqrt{\frac{2 h_{1}}{g}} a n d t_{2} = \sqrt{\frac{2 h_{2}}{g}}$
So, $\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$

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Two particles of masses m1andm2 are dropped from height h1andh2. They reach the earth after times t1andt2 respectively, Then:

The correct option is A t1t2=√h1h2h1=12gt21andh2=12gt22⇒t1=√2h1gandt2=√2h2gSo, t1t2=√h1h2

Two particles of masses $m_{1} a n d m_{2}$ are dropped from height $h_{1} a n d h_{2}$ . They reach the earth after times $t_{1} a n d t_{2}$ respectively, Then:

The correct option is A $\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$
$h_{1} = \frac{1}{2} g t_{1}^{2} a n d h_{2} = \frac{1}{2} g t_{2}^{2}$
$\Rightarrow t_{1} = \sqrt{\frac{2 h_{1}}{g}} a n d t_{2} = \sqrt{\frac{2 h_{2}}{g}}$
So, $\frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}}$