Question

A particle is projected vertically upwards. If it has to stay above the ground for 12 seconds, then
(A) velocity of projection is $192 \mathrm{ft} / \mathrm{sec}$
(B) greatest height attained is $600 \mathrm{ft}$
(C) velocity of projection is $196 \mathrm{ft} / \mathrm{sec}$
(D) greatest height attained is $576 \mathrm{ft}$

Answer 1

Ans : $(A, D)$
Hint : $V=u-g t$ at $t=6$
$\mathrm{u}-\mathrm{gt}=0$
$\Rightarrow \mathrm{u}=6 \mathrm{~g}=192 \mathrm{ft} / \mathrm{sec} \quad\left(\mathrm{g}=32 \mathrm{ft} / \mathrm{sec}^{2}\right) \ldots \ldots . .(\mathrm{i})$
$\mathrm{x}=\mathrm{ut}=\frac{1}{2} \mathrm{gt}^{2}$ $=192.6-\frac{1}{2} 32.6^{2}$
$=576 \mathrm{ft}$