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 dropped from top of a tower when it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of tower simultaneously. The height of tower is $(g=10 m/s^2)$

\[(a)\;90\; m \quad (b)\;45 \;m \quad (c)\;35 \;m \quad(d)\;25\; m \]

Can you answer this question?

1 Answer

0 votes
Let height of tower be 'h'
Time taken for 1st stone to reach ground is $=\sqrt {\large\frac{2h}{g}}$
$\bigg[s=ut+\large\frac{1}{2}$$gt^2$$; h=\large\frac{1}{2} $$gt^2\bigg]$
Time taken for the first stone to be travel 5 m from top $=\sqrt {\large\frac{2 \times 5}{g}}$
Time taken for 2nd stone to reach ground from a point $(h-25)m$ above the ground is $\sqrt {\large\frac{2(h-25)}{g}}$
Therefore $ \sqrt {\large\frac{2h}{g}}-\sqrt {\large\frac{2 \times 5}{g}}=\sqrt {\large\frac{2(h-25)}{g}}$
$\sqrt {2h}-\sqrt {10}=\sqrt {2(h-25)}$
Solving we get $h=45 \;m$
Hence b is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

Related questions

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