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 (g is acceleration due to gravity).

\frac{g \sqrt{t_{1} t_{2}}}{2}

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

\frac{g (t_{1} + t_{2})}{2}

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

g \sqrt{t_{1} t_{2}}

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

\frac{g t_{1} t_{2}}{t_{1} + t_{2}}

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

Open in App

Solution

The correct option is B $\frac{g (t_{1} + t_{2})}{2}$
When the body is projected, it takes time $t_{1}$ , to cross h, then another $t_{1}^{'}$ to go to $h_{m a x}$ when $v = 0$ . Then the initial velocity become $u^{'} = 0$ and takes time t to reach the ground.
$0 = u - g t$
$t - u / g$ .
$t_{1} + t_{1}^{'} = t = u / g$ .
$∴ t_{1} + t_{1}^{'} = u / g$ .
$t_{1} = (u / g) - t_{1}^{'}$
$t_{2} = t + t_{1}^{'} = \frac{u}{g} + t_{1}^{'}$
$∴ t_{1} + t_{2} = (2 u) / g$
$\Rightarrow u = \frac{g}{2} (t_{1} + t_{2})$ .

Suggest Corrections

Similar questions

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:

Q. 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 will be(take g as acceleration due to gravity).

Q. A particle is projected upwards. The times corresponding to height

h

while ascending and while descending are

t_{1}

and

t_{2}

respectively. The velocity of projection will be

Q. For a particle projected upwards to a height

h

, the ascending and descending times are

t_{1}

and

t_{2}

respectively. The velocity of projection will be

Q. A body is projected vertically upwards at time t = 0 and it is seen at a height H at time

t_{1}

and

t_{2}

second during its flight. The maximum height attained is (g is acceleration due to gravity)

Join BYJU'S Learning Program

A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are t1 and t2 respectively. Then the velocity of projection is (g is acceleration due to gravity).

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 (g is acceleration due to gravity).