Q107 / 155 · WBJEE 2020 · Mathematics
A particle is projected vertically upwards. If it has to stay above the ground for 12 seconds, then
- Correct answerAvelocity of projection is $192 \mathrm{ft} / \mathrm{sec}$
- Bgreatest height attained is $600 \mathrm{ft}$
- Cvelocity of projection is $196 \mathrm{ft} / \mathrm{sec}$
- Correct answerDgreatest height attained is $576 \mathrm{ft}$
Solution
Hint:
$V=u-g t$ at $t=6$
$u-g t=0$
$\Rightarrow u=6 g=192 f t /$ sec $\quad\left(g=32 f t / \sec ^{2}\right) \ldots \ldots . .(\mathrm{i})$
$x=u t=\frac{1}{2} g t^{2}$
$=192.6-\frac{1}{2} 32.6^{2}$
$=576 \mathrm{ft}$
Hence, the correct answer is options: 'A', 'D'.