The equations of motion of two stones thrown | vedclass

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(D) For the first stone,$s_1 = 19.6t - 4.9t^2$. 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

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(D) For the first stone,$s_1 = 19.6t - 4.9t^2$. 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 $t = 2 \, s$. The maximum height $h$ is $s_1(2) = 19.6(2) - 4.9(2^2) = 39.2 - 19.6 = 19.6 \, m$. Now,for the second stone,$s_2 = 9.8t - 4.9t^2$. At $t = 2 \, s$,the height of the second stone is $s_2(2) = 9.8(2) - 4.9(2^2) = 19.6 - 19.6 = 0 \, m$. Thus,when the first stone is at its maximum height,the second stone is on the ground.

The equations of motion of two stones thrown vertically upwards simultaneously are $s_1 = 19.6t - 4.9t^2$ and $s_2 = 9.8t - 4.9t^2$ respectively. If the maximum height attained by the first stone is $h$,then when the first stone is at its maximum height,the height of the second stone will be:

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