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A body is projected vertically upwards. If $ t_1$ and $t_2$ be the times at which it is at height h above the point of projection while ascending and descending respectively. then h is,

$\begin{array}{1 1}(A)\;\frac{1}{2}g t_1t_2 \\(B)\;g t_1t_2\\(C)\;2g t_1t_2\\(D)\;4g t_1t_2 \end{array} $

1 Answer

$h= ut - \large\frac{1}{2} $$gt^2$
If the roots of this equation are $t_1$ and $t_2$ then using the fact that the product of roots is equal to c/a for the equation $ax^2 +bx+c=0$ , we get
$t_1t_2= \large\frac{h}{(1/2g)}$
$\quad h= \large\frac{1}{2}$$gt_1t_2$
Hence A is the correct answer.
answered Jul 17, 2014 by meena.p

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