A body is projected up with velocity - u- It reaches the same -160-point in its path at - t-1- and - t-2- -160-from the time of -160-projection- Then - t-1-t-2- -160-is- 2ug- ug- -2ug- -ug-

The correct option is A 2ug When a body is projected up, suppose it crosses a point P at time t_1 #160;in its upward journey. The body will cross ...

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Byju's Answer

Standard XII

Physics

1st Equation of Motion

A body is pro...

Question

A body is projected up with velocity $u$ . It reaches the same point in its path at $t_{1}$ and $t_{2}$ from the time of projection. Then $t_{1} + t_{2}$ is:

\frac{2 u}{g}

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\frac{u}{g}

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\sqrt{\frac{2 u}{g}}

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\sqrt{\frac{u}{g}}

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Solution

The correct option is A $\frac{2 u}{g}$
When a body is projected up, suppose it crosses a point P at time $t_{1}$ in its upward journey. The body will cross the same point P at time $t_{2}$ in its downward journey.
Now the equation of trajectory is,
$h = u t - \frac{1}{2} g t^{2}$
Rearranging we get, $\frac{1}{2} g t^{2} - u t + h = 0$
These two values of ' $t$ ' give the value of $t_{1}$ and $t_{2}$
sum of the roots = $\frac{- b}{a} = \frac{u}{\frac{g}{2}} = \frac{2 u}{g}$
So, $t_{1} + t_{2} = \frac{2 u}{g}$

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