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| 1. |
Distance covered by ball thrown upwards with speed 'u' in last 't' seconds before it reaches max. height is ............. |
| Answer» <html><body><p>` (u + g t)t ` <br/>ut <br/>`1/<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> g t^(2)`<br/>`ut -(1)/(2) g t^(2)` </p>Solution :` (##KPK_AIO_PHY_XI_P1_C03_E06_001_S01.png" width="<a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a>%"> <br/> Suppose time taken by ball to reach max, height is T. <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> v = u - gT In v =0` <br/> Final velocity is zero at max. height <br/> `therefore u = gT` <br/> `therefore T = u/g ""…(1)`<br/> velocity gained by ball in `(T-t)s,` <br/> `v =u -g (T-t)` <br/> `=u -g ((u)/(g) -t) (because` From (1)) <br/> `=u - u + g t` <br/> `therefore v = g t""...(2)` <br/> Distance <a href="https://interviewquestions.tuteehub.com/tag/covered-2552266" style="font-weight:bold;" target="_blank" title="Click to know more about COVERED">COVERED</a> in last .t. <a href="https://interviewquestions.tuteehub.com/tag/seconds-1198593" style="font-weight:bold;" target="_blank" title="Click to know more about SECONDS">SECONDS</a>, <br/> `d = vt - 1/2 g t^(2)` <br/> `= (g t)t - 1/2 g t^(2) (because ` For (2)) <br/> `= g t ^(2) -1/2 g t^(2)` <br/> `therefore d = 1/2 g t^(2)`</body></html> | |