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