Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
$-h= ut_1- \frac{1}{2}gt_1^2$--------(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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App