Question

A stone was thrown down with speed it takes a time to reach the ground while another stone thrown upwards from the same point with the same speed takes time t2 the maximum height the second stone reaches from the ground is:

Answer 1

Case of t1 the stone was thrown downward.
$H=ut1+\frac{1}{2}gt{1}^2$--(a)

second case $-H=ut2+\frac{1}{2}gt{2}^2$ ..(b)

Maximum height with respect to the level H can be calculated by $v2=u2-2as=>s=\frac{u^2}{2g}$ so with height will be $\frac{H+u^2}{2g}$ But $H$ is not given So by subtraction a and b equations

$2H=(ut1+\frac{1}{2}gt{1}^2)–(ut2+\frac{1}{2}gt{2}^2)$

$=u(t1-t2)-\left(\frac{1}{2}gt{1}^2\right)-t{2}^2$

$H=\frac{u(t1-t2)-\frac{1}{2}g(t{1}^2-t{2}^2)}{2}$
so the maxium height with respect to groun is $=\frac{u(t1-t2)-\frac{1}{2}g(t{1}^2-t{2}^2)}{2}+\frac{u^2}{2g}$