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A body projected from the ground reaches a point $X$ in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point $X$ from the ground is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$90\ m$
Total time of flight $(t)=\frac{2 u}{g}$
or, $9=\frac{2 u}{g} \quad$ or, $u=\frac{9 g}{2}$
or, $\quad u=\frac{9 \times 10}{2}$ or, $u=45 \mathrm{~m} / \mathrm{s}$
Since, in covering the vertical distance, $g$ becomes - (ve)
$\begin{aligned}
& \text { using } h=u t-\frac{1}{2} g t^2 \\
& =45 \times(3)-\frac{1}{2} \times 10 \times(3)^2=135-45=90 \mathrm{~m}
\end{aligned}$
or, $9=\frac{2 u}{g} \quad$ or, $u=\frac{9 g}{2}$
or, $\quad u=\frac{9 \times 10}{2}$ or, $u=45 \mathrm{~m} / \mathrm{s}$
Since, in covering the vertical distance, $g$ becomes - (ve)
$\begin{aligned}
& \text { using } h=u t-\frac{1}{2} g t^2 \\
& =45 \times(3)-\frac{1}{2} \times 10 \times(3)^2=135-45=90 \mathrm{~m}
\end{aligned}$
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