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The height reached in time t by a particle thrown upwared with a speed u is given by `h=ut-1/2 gt^2`where g=9.8 m/s^2 is a constant.Find the timetakenin raching the maximum height. |
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Answer» The height h is a function of tiem.Thus, h will be maximum when `(dh)/(dt)=0` We have, `h=ut-1/2 gt^2` or, `(dh)/(dt)=d/(dt) (ut)-d/dt(1/2 gt^2)` `=u(dt)/(dt)-/2g d/(dt) (t^2)` `= u-1/2 g(2g)=u-gt.` For maximum h, `(dh)/(dt)=0` or, `u-gt=0 or, t=u/g` |
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