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

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

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

$t_{2} - t_{1} = \frac{2 u}{g}$

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

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

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

$t_{2} - t_{1} = \frac{u}{g}$

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

Open in App

Solution

The correct option is A
$t_{1} + t_{2} = \frac{2 u}{g}$

From the second equation of motion, $h = u t - \frac{1}{2} g t^{2}$
$\Rightarrow t^{2} - \frac{2 u}{g} t + \frac{2 h}{g} = 0$
The roots of this quadratic equation are $t_{1}$ and $t_{2}$ .
The sum of the roots is $t_{1} + t_{2} = \frac{2 u}{g}$ . Hence, the correct option is (a).

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

Q. 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:

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: