The time of flight of a vertically projected stone is $8 \mathrm{~s}$. The position of the stone after $6…
The time of flight of a vertically projected stone is $8 \mathrm{~s}$. The position of the stone after $6 \mathrm{~s}$ from the ground is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$20 \mathrm{~m}$
$60 \mathrm{~m}$
$75 \mathrm{~m}$
$40 \mathrm{~m}$
Solution
Time of flight $\mathrm{t}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$ or, $8=\frac{2 \times \mathrm{u} \times 1}{10}$ $\Rightarrow \mathrm{u}=40 \mathrm{~m} / \mathrm{s}$
$\therefore$ The position of the stone after $6 \mathrm{~s}$ from the ground.
$
\begin{aligned}
& \mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2=40 \times 6+\frac{1}{2}(-10) \times 6^2 \\
& =240-180=60 \mathrm{~m}
\end{aligned}
$