Question

A body is projected vertically upwards. The time corresponding to height $h$ while ascending and while descending are ${t_1}$ and ${t_2}$ respectively. Then the velocity of projection is ($g$ is the acceleration due to gravity.)
$(A)\dfrac{{g\sqrt {{t_1}{t_2}} }}{2}$
$(B)\dfrac{{g\sqrt {{t_1} + {t_2}} }}{2}$
$(C)g\sqrt {{t_1}{t_2}}$
$(D)\dfrac{{g{t_1}{t_2}}}{{{t_1} + {t_2}}}$

Answer 1

The velocity of projection is defined as the velocity with which the body is thrown. The point from which the body is projected in the air is defined as a point of projection. We need to use the second equation of motion to find the times at which the body will be the height with the time as the unknown variable and the second equation of motion is a quadratic equation.
Formula Used:
The given formula is used to find the velocity of the projection.
$h = ut + \dfrac{1}{2}g{t^2}$
Where,
$h$-is denoted as the height of an object thrown vertically upward
$u$-is the initial velocity of the projection
$g$-is the acceleration due to gravity
$t$-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 answer:
When the 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
$h = ut + \dfrac{1}{2}g{t^2}$
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
$t = \dfrac{{u - \sqrt {{u^2} - 2gH} }}{g}$
And
$t = \dfrac{{u + \sqrt {{u^2} - 2gH} }}{g}$
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 answer is the velocity of projection is $u = \dfrac{{g\left( {{t_1} + {t_2}} \right)}}{2}$.

Note: The equation $H = ut - \dfrac{1}{2}g{t^2}$ has been derived by allowing downward to be negative.
Hence the upward can be taken as the negative instead.
When upward is negative the height $h = - H$ and not $H$, since the height is measured upward from the ground.