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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Time taken for the particle to reach highest point is $ \large\frac{t_1+t_2}{2}$

$(v=u-gt)$

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$

$h=ut_1-\large\frac{1}{2}$$g{t_1}^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.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...