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A particle is projected vertically upwards. If it has to stay above the ground for 12 seconds, then
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Verified Answer
The correct answers are:
velocity of projection is $192 \mathrm{ft} / \mathrm{sec}$, greatest height attained is $576 \mathrm{ft}$
Hint:
$V=u-g t$ at $t=6$
$u-g t=0$
$\Rightarrow u=6 g=192 f t /$ sec $\quad\left(g=32 f t / \sec ^{2}\right) \ldots \ldots . .(\mathrm{i})$
$x=u t=\frac{1}{2} g t^{2}$
$=192.6-\frac{1}{2} 32.6^{2}$
$=576 \mathrm{ft}$
$V=u-g t$ at $t=6$
$u-g t=0$
$\Rightarrow u=6 g=192 f t /$ sec $\quad\left(g=32 f t / \sec ^{2}\right) \ldots \ldots . .(\mathrm{i})$
$x=u t=\frac{1}{2} g t^{2}$
$=192.6-\frac{1}{2} 32.6^{2}$
$=576 \mathrm{ft}$
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