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(D) For the first stone,the equation of motion is $s_1 = 19.6t - 4.9t^2$. The velocity is $v_1 = \frac{ds_1}{dt} = 19.6 - 9.8t$. At maximum height,$v_

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(D) For the first stone,the equation of motion is $s_1 = 19.6t - 4.9t^2$. The velocity is $v_1 = \frac{ds_1}{dt} = 19.6 - 9.8t$. At maximum height,$v_1 = 0$,so $19.6 - 9.8t = 0$,which gives $t = 2 	ext{ s}$.The maximum height $h$ reached by the first stone is $s_1(2) = 19.6(2) - 4.9(2)^2 = 39.2 - 19.6 = 19.6 	ext{ m}$.Now,consider the second stone with equation $s_2 = 9.8t - 4.9t^2$. The velocity is $v_2 = \frac{ds_2}{dt} = 9.8 - 9.8t$. The second stone reaches its maximum height at $t = 1 	ext{ s}$,where $s_2(1) = 9.8(1) - 4.9(1)^2 = 4.9 	ext{ m}$.After $t = 1 	ext{ s}$,the second stone starts moving downwards and hits the ground when $s_2 = 0$,i.e.,$9.8t - 4.9t^2 = 0 \implies 4.9t(2 - t) = 0$,so $t = 2 	ext{ s}$.At $t = 2 	ext{ s}$,the first stone is at its maximum height $h$,and the second stone is at $s_2(2) = 9.8(2) - 4.9(2)^2 = 19.6 - 19.6 = 0 	ext{ m}$.

Two stones are thrown vertically upwards with equations of motion $s_1 = 19.6t - 4.9t^2$ and $s_2 = 9.8t - 4.9t^2$ respectively. The maximum height of the first stone is $h$. When the first stone is at its maximum height,the height of the second stone is:

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