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| 1. |
Ratio of distances covered by object falling freely under gravity in 1 ^(st),2^(nd)&3^(rd)second 18 ........ |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>:3:5`<br/>`1:2:5`<br/>`1:4:9`<br/>`1:5:9`</p>Solution :Distance covered in `n^(<a href="https://interviewquestions.tuteehub.com/tag/th-665760" style="font-weight:bold;" target="_blank" title="Click to know more about TH">TH</a>)`second, <br/> In `d _(n) = v _(0)t + a/2 (2n -1) , v_(0) =0,` <br/> `d_(n) = (a)/(2) (2n -1)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> ` For `n =1, d _(1) = g /2 (<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a> 1-1) = g/2` <br/> For `n = 2, d_(2) = g/2 (2 xx 2-1) = (3g)/(2)` <br/> For `n =3, d _(3) = g/2) (2 xx 3-1) = (<a href="https://interviewquestions.tuteehub.com/tag/5g-326287" style="font-weight:bold;" target="_blank" title="Click to know more about 5G">5G</a>)/(2)` <br/> `therefore d _(1) : d_(2) :d _(3) = (g)/(2) : (3g)/(2) : (5g)/(2) = 1:3:5`</body></html> | |