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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{u}{g}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $t_{1} + t_{2} = \frac{2 u}{g}$

Suggest Corrections

Similar questions

A body is thrown vertically upwards with an initial velocity u. It passes a point at a height h above the ground at time $t_{1}$ while going up and at time $t_{2}$ while falling down. Then the relation between u, $t_{1}$ and $t_{2}$ is

A body is projected upward with a velocity u. It passes through a certain point above the ground after $t_{1}$ . From this instant, the time after which the body passes through the same point during the return journey is

Q. A particle is projected with speed

u

at angle

θ

with horizontal from ground. If it is at same height from ground at time

t_{1}

and

t_{2}

, then its average velocity in time interval

t_{1}

t_{2}

A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then, the velocity of projection is:

Join BYJU'S Learning Program

A body is projected up with velocity u. It reaches the same point in its path at t1 and t2 from the time of projection. Then t1+t2 is:

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: