Question

A particle is projected vertically upwards. If it has to stay above the ground for $12$ seconds, then

• A. velocity of projection is $192\text{ft/sec}$
• B. greatest height attained is $600\text{ft}$
• C. velocity of projection is $196\text{ft/sec}$
• D. greatest height attained is $576\text{ft}$

Answer 1

Answer: (A), (D)

The correct options are
A velocity of projection is $192\text{ft/sec}$
D greatest height attained is $576\text{ft}$

$v=u-gt$
$v=0,g=32\text{ft/sec},t=6\text{sec}$
$\Rightarrow 0=u-32\times 6$
$\Rightarrow u=192\text{ft/sec}$

$h=ut-\frac{1}{2}g{t}^2$
$\Rightarrow h=192\times 6-\frac{1}{2}\times 32\times 6^2$
$=576\text{ft}$