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A stone thrown down with a speed $u$ takes a time $t_1$ to reach the ground, while another stone thrown upwards from the same point with the same speed takes time $t_2$. The maximum height the second stone reaches from the ground is
$\frac{1}{2} g t_1 t_2$
$g / 8\left(t_1+t_2\right)^2$
$g / 8\left(t_1-t_2\right)^2$
$\frac{1}{2} g t_2^2$
Solution
(b)
$-h=-u t_1+\frac{1}{2}(-g) t_1^2$
or
$h=u t_1+\frac{1}{2} g t_1^2 \quad \dots(i)$
$-h=u t_2-\frac{1}{2} g t_2^2 \quad \dots(ii)$
Adding Eqs. $(i)$ and $(ii)$, we get
$0=u\left(t_1+t_2\right)-\frac{1}{2} g\left(t_1^2-t_2^2\right)$
$\Rightarrow \quad u=\frac{g}{2}\left(t_1-t_2\right)$
Maximum height attained by second stone is
$H=h+\frac{u^2}{2 g}$
$\Rightarrow \quad H=u t_1+\frac{1}{2} g t_1^2+\frac{u^2}{2 g}$
Substituting for $u$ and rearranging,
we get
$H=\frac{g}{8}\left(t_1+t_2\right)^2$
