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A stone is thrown vertically up from the top of a tower with some initial velocity and it arrives on the ground after $t_1$ seconds. Now if the same the stone is thrown vertically down from the top of the same tower with the same initial velocity, it arrives on ground after $t_2$ seconds. In how much time will the stone take to reach the ground if it is dropped from the same tower?

$\begin{array}{1 1}(A)\;\frac{t_1+t_2}{2} \\(B)\;\frac{t_1-t_2}{2}\\(C)\;t_1+t_2\\(D)\;\sqrt {t_1t_2} \end{array} $

1 Answer

$-h= ut_1- \frac{1}{2}gt_1^2$--------(i)
$h= ut_2+\large\frac{1}{2}$$gt_2^2$--------(ii)
Where h is a positive number. Multiplying (i) by $t_2$ and (ii) by $t_1$ we get,
$h(t_1+t_2)=\large\frac{1}{2}=\large\frac{1}{2} $$gt_1t_2(t_1+t_2)$-----------(iii)
$y= \large\frac{1}{2} $$gt^2$---------(iv)
From (iii) and (iv)
$t= \sqrt {t_1t_2}$
Hence D is the correct answer.
answered Jul 17, 2014 by meena.p

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