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# 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)(A) $\dfrac{{g\sqrt {{t_1}{t_2}} }}{2}$(B) $\dfrac{{g\left( {{t_1} + {t_2}} \right)}}{2}$(C) $g\sqrt {{t_1}{t_2}}$(D) $g\dfrac{{{t_1}{t_2}}}{{\left( {{t_1} + {t_2}} \right)}}$

Last updated date: 29th Feb 2024
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Hint: We need to use the second equation of motion to find the times at which the body will be at height $H$. With time as the unknown variable, the second equation of motion is a quadratic equation.
Formula used: In this solution we will be using the following formulae;
$h = ut + \dfrac{1}{2}g{t^2}$ where $h$ is the height of an object thrown vertically upward, $u$ is the initial velocity of projection, $g$ is the acceleration due to gravity, and $t$ is the instantaneous time when at the height.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for an equation $a{x^2} + bx + c = 0$

Complete Step-by-Step solution:
Generally, when a body is thrown upwards, any particular height is crossed twice, while going up and while coming down. The second equation of motion can be used to calculate the times when it crosses a particular height. This is given as
$h = ut + \dfrac{1}{2}g{t^2}$ where $h$ is the height of an object thrown vertically upward, $u$ is the initial velocity of projection, $g$ is the acceleration due to gravity, and $t$ is the instantaneous time when at the height.
By inserting known values, we have
$H = ut - \dfrac{1}{2}g{t^2}$
$\Rightarrow g{t^2} - 2ut + 2H = 0$ which is a quadratic equation.
Hence, using the quadratic formula given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for an equation $a{x^2} + bx + c = 0$.
Hence, we have
$t = \dfrac{{2u \pm \sqrt {{{\left( {2u} \right)}^2} - 4\left( g \right)\left( {2H} \right)} }}{{2g}} = \dfrac{{2u \pm \sqrt {4{u^2} - 8gH} }}{{2g}}$
By simplification, we have
$t = \dfrac{{u \pm \sqrt {{u^2} - 2gH} }}{g}$
Then the two different values of $t$ are
${t_1} = \dfrac{{u - \sqrt {{u^2} - 2gH} }}{g}$
and
${t_2} = \dfrac{{u + \sqrt {{u^2} - 2gH} }}{g}$
By adding the two times, we have
${t_1} + {t_2} = \dfrac{{u + \sqrt {{u^2} - 2gH} }}{g} + \dfrac{{u - \sqrt {{u^2} - 2gH} }}{g} = \dfrac{{2u}}{g}$
By making $u$ subject, we have
$u = \dfrac{{g\left( {{t_1} + {t_2}} \right)}}{2}$

Hence, the correct option is B

Note: For clarity, the equation $H = ut - \dfrac{1}{2}g{t^2}$ has been derived by allowing downward to be negative. This is a matter of choice, and hence upward can be taken as negative instead. However, note that when upward is negative the height $h = - H$, and not $H$, since the height is measured upward from the ground.