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A body is projected vertically upwards. The times corresponding to height while ascending and while descending are $\mathrm{t}_{1}$ and $\mathrm{t}_{2}$ respectively. Then the velocity of projection is ( is acceleration due to gravity)
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The correct answer is:
$\frac{g\left(t_{1}+t_{2}\right)}{2}$
In case of motion under gravity, time taken to go up is equal to the time taken to fall down through the same distance.
Time of descent $\left(t_{2}\right)=$ time of ascent $\left(t_{1}\right)=\frac{u}{g}$
$\therefore$ Total time of flight $\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{2 \mathrm{u}}{\mathrm{g}}$
$\Rightarrow \quad \mathrm{u}=\frac{g\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}{2}$
Time of descent $\left(t_{2}\right)=$ time of ascent $\left(t_{1}\right)=\frac{u}{g}$
$\therefore$ Total time of flight $\mathrm{T}=\mathrm{t}_{1}+\mathrm{t}_{2}=\frac{2 \mathrm{u}}{\mathrm{g}}$
$\Rightarrow \quad \mathrm{u}=\frac{g\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}{2}$
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