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(B) Let $h$ be the height of the point from the ground.For the first stone thrown downwards with speed $u$:$-h = -u t_1 - \frac{1}{2} g t_1^2 \Rightar

Vedclass · Accepted Answer

$\frac{g}{8} (t_1 + t_2)^2$

Vedclass · Answer

(B) Let $h$ be the height of the point from the ground.For the first stone thrown downwards with speed $u$:$-h = -u t_1 - \frac{1}{2} g t_1^2 \Rightarrow h = u t_1 + \frac{1}{2} g t_1^2 \quad \dots(i)$For the second stone thrown upwards with speed $u$:$-h = u t_2 - \frac{1}{2} g t_2^2 \Rightarrow h = \frac{1}{2} g t_2^2 - u t_2 \quad \dots(ii)$Equating $(i)$ and $(ii)$:$u t_1 + \frac{1}{2} g t_1^2 = \frac{1}{2} g t_2^2 - u t_2$$u(t_1 + t_2) = \frac{1}{2} g (t_2^2 - t_1^2) = \frac{1}{2} g (t_2 - t_1)(t_2 + t_1)$$u = \frac{g}{2} (t_2 - t_1)$Maximum height $H$ reached by the second stone from the ground is the height of the point $h$ plus the height reached above the point:$H = h + \frac{u^2}{2g}$Substituting $h$ from $(ii)$ and $u$:$H = (\frac{1}{2} g t_2^2 - u t_2) + \frac{u^2}{2g}$$H = \frac{1}{2} g t_2^2 - \frac{g}{2}(t_2 - t_1)t_2 + \frac{1}{2g} \cdot \frac{g^2}{4}(t_2 - t_1)^2$$H = \frac{1}{2} g t_2^2 - \frac{1}{2} g t_2^2 + \frac{1}{2} g t_1 t_2 + \frac{g}{8}(t_2^2 - 2 t_1 t_2 + t_1^2)$$H = \frac{g}{8}(4 t_1 t_2 + t_2^2 - 2 t_1 t_2 + t_1^2) = \frac{g}{8}(t_1^2 + 2 t_1 t_2 + t_2^2) = \frac{g}{8}(t_1 + t_2)^2$

$A$ 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

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