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A stone is dropped from the top of a cliff and 'n' second later, another stone is thrown vertically downwards, with a velocity V. How far below the top of

the cliff will the second stone overtake the first?




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Let the second stone overtakes the first stone t seconds after the first stone is dropped.

height covered by first stone = 0.5gt²
height covered by second stone = v(t-n)+0.5g(t-n)²
Both heights are equal. Thus
0.5gt² = v(t-n)+0.5g(t-n)²
⇒0.5gt² = vt - vn + 0.5g(t²-2tn+n²)
⇒0.5gt² = vt - vn + 0.5gt² - gtn + 0.5gn²
⇒vt - vn - gtn + 0.5gn² = 0
⇒t(v-gn) - vn + 0.5gn² = 0
⇒t(v-gn) = vn - 0.5gn²

⇒ t =  \frac{vn-0.5gn^{2} }{v-gn}

Distance travelled = 0.5gt² = 0.5g [\frac{vn-0.5gn^{2} }{v-gn}  ]^{2}
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