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 particle is projected vertically from a point A on ground. It takes time $t_1$ to reach point B but still continous to move up. It takes further $t_2$ time to reach the ground from point B. Height of point B from the ground is

\[(a)\;\frac{1}{2} g(t_1+t_2)^2\quad (b)\;g t_1t_2 \quad (c)\;\frac{1}{8} g(t_1+t_2)^2 \quad (d)\frac{1}{2} g t_1t_2\]

Can you answer this question?

1 Answer

0 votes
Time taken for the particle to reach highest point is $ \large\frac{t_1+t_2}{2}$
Therefore initial velocity$=u=g \bigg(\frac{t_1+t_2}{2}\bigg)$
Therefore heigh of B from ground is
$h=ut+\large\frac{1}{2}$$at ^2$
$\quad=g \bigg(\large\frac{t_1+t_2}{2}\bigg)$$t_1-\large\frac{1}{2} $$gt_1^2$
$h=\large\frac{1}{2}$$g t_1 t_2$
Hence d is the correct answer.


answered Jun 26, 2013 by meena.p
edited Jan 25, 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