A dynamite blast blows a heavy rock straight up with a launch velocity of $160 m / \sec .$ It reaches a height of $S = 160 t - 16 t^{2}$ after t sec. The velocity of rock when it is $256 m$ above the ground on the way up is V m/sec then $\frac{V}{12}$ is _______

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answer is 8.

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Detailed Solution

$V = \frac{d S}{d t} = 160 - 32 t \Rightarrow S (t) = 256 \Rightarrow 160 t - 16 t^{2} = 256 \Rightarrow t = 2 o r 8 \Rightarrow V (2) = 160 - 64 = 96 m / \sec \Rightarrow \frac{V}{12} = \frac{96}{12} = 8$

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