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} $ .