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(C) Given the displacement from the top of the tower is $s=48t-16t^{2}$.The velocity is given by $v = \frac{ds}{dt} = 48-32t$.At the greatest height,t

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$100$

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(C) Given the displacement from the top of the tower is $s=48t-16t^{2}$.The velocity is given by $v = \frac{ds}{dt} = 48-32t$.At the greatest height,the velocity $v=0$.$48-32t=0 \Rightarrow t = \frac{48}{32} = 1.5 \ s$.The maximum displacement from the top of the tower is $s = 48(1.5) - 16(1.5)^{2} = 72 - 36 = 36 \ m$.The total height above the ground is the height of the tower plus the maximum displacement: $H = 64 + 36 = 100 \ m$.

$A$ stone is thrown vertically upwards from the top of a tower $64 \ m$ high according to the law $s=48t-16t^{2}$. The greatest height attained by the stone above the ground is (in $m$)

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