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The height reached in time t by a particle thrown upward with a speed u is given by `h=ut-(1)/(2) "gt"^(2)` where g is acceleration due to gravity. Find the time taken in reaching the maximum height. |
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Answer» Here h=f(t). Thus h will be maximum when `(dh)/(dt)=0` We have `h=ut- (1)/(2) "gt"^(2)` or `(dh)/(dt)=u-(1)/(2)g(2t)=u-"gt"` for maximum h, `(dh)/(dt)=0` `rArr " " u-"gt"=0` `t=(u)/(g)` |
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