Question

A particle is projected vertically upwards from a point A on the ground. It takes time ${t_1}$ , to reach a point B, but it still continues to move up. It takes further time ${t_2}$ to reach the ground from point B. the height of point B from the ground is-

Answer 1

Hint : The total time of flight is equal to the sum of the time to get to point B and the time to go from point B back to the ground. The final height from the ground should only be expressed in terms of all known values (i.e. constants and variables given)

Formula used: In this solution we will be using the following formula;
$h = ut \pm \dfrac{1}{2}g{t^2}$ where $h$ is the height of an upwardly projected object above the ground, $u$ is the initial velocity of projection, $t$ is the particular considered, and $g$ is the acceleration due to gravity.
$T = \dfrac{{2u}}{g}$ where $T$ is the time of flight.

Complete step by step answer
A particular particle is projected vertically upwards from point A on the ground. It is said to reach point B after a time ${t_1}$ , we are to calculate the height of B above the ground.
The height above the ground of such motion is given as
$h = ut \pm \dfrac{1}{2}g{t^2}$ where $u$ is the initial velocity of projection, $t$ is the particular consideration, and $g$ is the acceleration due to gravity. Hence, for our particle at point B it is
${h_B} = u{t_1} - \dfrac{1}{2}g{t_1}^2$ ( on assuming downward is negative)
All variables, except the initial velocity of projection $u$ , is known. Hence, we need to find $u$ .
The time of flight (time taken to complete whole journey) is given as
$T = \dfrac{{2u}}{g}$ , hence, making $u$ subject of formula, we have that
$u = \dfrac{{Tg}}{2}$
But according to question, it takes a time ${t_2}$ to go from point B back to the ground, hence, the total time of flight is
$T = {t_1} + {t_2}$ , thus by insertion,
$u = \dfrac{{g({t_1} + {t_2})}}{2}$ , then finally, substituting into ${h_B} = u{t_1} + \dfrac{1}{2}g{t_1}^2$ , we have
${h_B} = \dfrac{{g({t_1} + {t_2})}}{2}{t_1} - \dfrac{1}{2}g{t_1}^2$
By simplifying
${h_B} = \dfrac{{g{t_1}}}{2}\left[ {({t_1} + {t_2}) - {t_1}} \right]$
$\Rightarrow {h_B} = \dfrac{{g{t_1}}}{2}\left( {{t_2}} \right)$
Thus,
$\therefore {h_B} = \dfrac{{g{t_1}{t_2}}}{2}$ .

Note
For clarity, the time of flight can be derived from the equation of the velocity
$v = u - gt$ where $v$ is the final velocity at a time $t$ . Hence, the for time of flight $t = T$ , the final velocity is equal to negative of the initial velocity (since they are in opposite direction) i.e. $v = - u$
Hence, $- u = u - gT$
$\Rightarrow 2u = gT$
By dividing both sides by $g$ , we have
$T = \dfrac{{2u}}{g}$ .