A body is projected with a velocity u. It passes through a certain point above the ground after t1 sec. The time after which the body passes through the same point during the return journey is

Let v be the velocity attained by the body after t1 sec.Then v = u -gt1We know that time taken to the reach the same pointt2=2vg=2gu−gt1or t2=2ug−t1

A body is projected with a velocity u. It passes through a certain point above the ground after t1 sec. The time after which the body passes through the same point during the return journey is

Let v be the velocity attained by the body after t1 sec.Then v = u -gt1We know that time taken to the reach the same pointt2=2vg=2gu−gt1or t2=2ug−t1

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A body is projected with a velocity u. It passes through a certain point above the ground after t₁ sec. The time after which the body passes through the same point during the return journey is

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$(\frac{u}{g} - t_{1}^{2})$

$2 (\frac{u}{g} - t_{1})$

$3 (\frac{u^{2}}{g} - t_{1})$

$3 (\frac{u^{2}}{g^{2}} - t_{1})$

answer is B.

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Detailed Solution

Let v be the velocity attained by the body after t₁ sec.
Then v = u -gt₁
We know that time taken to the reach the same point
$\begin{array}{l} t_{2} = \frac{2 v}{g} = \frac{2}{g} (u - {gt}_{1}) \\ or t_{2} = 2 (\frac{u}{g} - t_{1}) \end{array}$

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A body is projected with a velocity u. It passes through a certain point above the ground after t1 sec. The time after which the body passes through the same point during the return journey is

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A body is projected with a velocity u. It passes through a certain point above the ground after t₁ sec. The time after which the body passes through the same point during the return journey is